--
-- JOIN
-- Test JOIN clauses
--

CREATE TABLE J1_TBL (
  i integer,
  j integer,
  t text
);

CREATE TABLE J2_TBL (
  i integer,
  k integer
);


INSERT INTO J1_TBL VALUES (1, 4, 'one');
INSERT INTO J1_TBL VALUES (2, 3, 'two');
INSERT INTO J1_TBL VALUES (3, 2, 'three');
INSERT INTO J1_TBL VALUES (4, 1, 'four');
INSERT INTO J1_TBL VALUES (5, 0, 'five');
INSERT INTO J1_TBL VALUES (6, 6, 'six');
INSERT INTO J1_TBL VALUES (7, 7, 'seven');
INSERT INTO J1_TBL VALUES (8, 8, 'eight');
INSERT INTO J1_TBL VALUES (0, NULL, 'zero');
INSERT INTO J1_TBL VALUES (NULL, NULL, 'null');
INSERT INTO J1_TBL VALUES (NULL, 0, 'zero');

INSERT INTO J2_TBL VALUES (1, -1);
INSERT INTO J2_TBL VALUES (2, 2);
INSERT INTO J2_TBL VALUES (3, -3);
INSERT INTO J2_TBL VALUES (2, 4);
INSERT INTO J2_TBL VALUES (5, -5);
INSERT INTO J2_TBL VALUES (5, -5);
INSERT INTO J2_TBL VALUES (0, NULL);
INSERT INTO J2_TBL VALUES (NULL, NULL);
INSERT INTO J2_TBL VALUES (NULL, 0);

--
-- CORRELATION NAMES
-- Make sure that table/column aliases are supported
-- before diving into more complex join syntax.
--

SELECT '' AS "xxx", *
  FROM J1_TBL AS tx;

SELECT '' AS "xxx", *
  FROM J1_TBL tx;

SELECT '' AS "xxx", *
  FROM J1_TBL AS t1 (a, b, c);

SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b, c);

SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b, c), J2_TBL t2 (d, e);

SELECT '' AS "xxx", t1.a, t2.e
  FROM J1_TBL t1 (a, b, c), J2_TBL t2 (d, e)
  WHERE t1.a = t2.d;


--
-- CROSS JOIN
-- Qualifications are not allowed on cross joins,
-- which degenerate into a standard unqualified inner join.
--

SELECT '' AS "xxx", *
  FROM J1_TBL CROSS JOIN J2_TBL;

-- ambiguous column
SELECT '' AS "xxx", i, k, t
  FROM J1_TBL CROSS JOIN J2_TBL;

-- resolve previous ambiguity by specifying the table name
SELECT '' AS "xxx", t1.i, k, t
  FROM J1_TBL t1 CROSS JOIN J2_TBL t2;

SELECT '' AS "xxx", ii, tt, kk
  FROM (J1_TBL CROSS JOIN J2_TBL)
    AS tx (ii, jj, tt, ii2, kk);

SELECT '' AS "xxx", tx.ii, tx.jj, tx.kk
  FROM (J1_TBL t1 (a, b, c) CROSS JOIN J2_TBL t2 (d, e))
    AS tx (ii, jj, tt, ii2, kk);

SELECT '' AS "xxx", *
  FROM J1_TBL CROSS JOIN J2_TBL a CROSS JOIN J2_TBL b;


--
--
-- Inner joins (equi-joins)
--
--

--
-- Inner joins (equi-joins) with USING clause
-- The USING syntax changes the shape of the resulting table
-- by including a column in the USING clause only once in the result.
--

-- Inner equi-join on specified column
SELECT '' AS "xxx", *
  FROM J1_TBL INNER JOIN J2_TBL USING (i);

-- Same as above, slightly different syntax
SELECT '' AS "xxx", *
  FROM J1_TBL JOIN J2_TBL USING (i);

SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b, c) JOIN J2_TBL t2 (a, d) USING (a)
  ORDER BY a, d;

SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b, c) JOIN J2_TBL t2 (a, b) USING (b)
  ORDER BY b, t1.a;


--
-- NATURAL JOIN
-- Inner equi-join on all columns with the same name
--

SELECT '' AS "xxx", *
  FROM J1_TBL NATURAL JOIN J2_TBL;

SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b, c) NATURAL JOIN J2_TBL t2 (a, d);

SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b, c) NATURAL JOIN J2_TBL t2 (d, a);

-- mismatch number of columns
-- currently, Postgres will fill in with underlying names
SELECT '' AS "xxx", *
  FROM J1_TBL t1 (a, b) NATURAL JOIN J2_TBL t2 (a);


--
-- Inner joins (equi-joins)
--

SELECT '' AS "xxx", *
  FROM J1_TBL JOIN J2_TBL ON (J1_TBL.i = J2_TBL.i);

SELECT '' AS "xxx", *
  FROM J1_TBL JOIN J2_TBL ON (J1_TBL.i = J2_TBL.k);


--
-- Non-equi-joins
--

SELECT '' AS "xxx", *
  FROM J1_TBL JOIN J2_TBL ON (J1_TBL.i <= J2_TBL.k);


--
-- Outer joins
-- Note that OUTER is a noise word
--

SELECT '' AS "xxx", *
  FROM J1_TBL LEFT OUTER JOIN J2_TBL USING (i)
  ORDER BY i, k, t;

SELECT '' AS "xxx", *
  FROM J1_TBL LEFT JOIN J2_TBL USING (i)
  ORDER BY i, k, t;

SELECT '' AS "xxx", *
  FROM J1_TBL RIGHT OUTER JOIN J2_TBL USING (i);

SELECT '' AS "xxx", *
  FROM J1_TBL RIGHT JOIN J2_TBL USING (i);

SELECT '' AS "xxx", *
  FROM J1_TBL FULL OUTER JOIN J2_TBL USING (i)
  ORDER BY i, k, t;

SELECT '' AS "xxx", *
  FROM J1_TBL FULL JOIN J2_TBL USING (i)
  ORDER BY i, k, t;

SELECT '' AS "xxx", *
  FROM J1_TBL LEFT JOIN J2_TBL USING (i) WHERE (k = 1);

SELECT '' AS "xxx", *
  FROM J1_TBL LEFT JOIN J2_TBL USING (i) WHERE (i = 1);


--
-- More complicated constructs
--

--
-- Multiway full join
--

CREATE TABLE t1 (name TEXT, n INTEGER);
CREATE TABLE t2 (name TEXT, n INTEGER);
CREATE TABLE t3 (name TEXT, n INTEGER);

INSERT INTO t1 VALUES ( 'bb', 11 );
INSERT INTO t2 VALUES ( 'bb', 12 );
INSERT INTO t2 VALUES ( 'cc', 22 );
INSERT INTO t2 VALUES ( 'ee', 42 );
INSERT INTO t3 VALUES ( 'bb', 13 );
INSERT INTO t3 VALUES ( 'cc', 23 );
INSERT INTO t3 VALUES ( 'dd', 33 );

SELECT * FROM t1 FULL JOIN t2 USING (name) FULL JOIN t3 USING (name);

--
-- Test interactions of join syntax and subqueries
--

-- Basic cases (we expect planner to pull up the subquery here)
SELECT * FROM
(SELECT * FROM t2) as s2
INNER JOIN
(SELECT * FROM t3) s3
USING (name);

SELECT * FROM
(SELECT * FROM t2) as s2
LEFT JOIN
(SELECT * FROM t3) s3
USING (name);

SELECT * FROM
(SELECT * FROM t2) as s2
FULL JOIN
(SELECT * FROM t3) s3
USING (name);

-- Cases with non-nullable expressions in subquery results;
-- make sure these go to null as expected
SELECT * FROM
(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2
NATURAL INNER JOIN
(SELECT name, n as s3_n, 3 as s3_2 FROM t3) s3;

SELECT * FROM
(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2
NATURAL LEFT JOIN
(SELECT name, n as s3_n, 3 as s3_2 FROM t3) s3;

SELECT * FROM
(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2
NATURAL FULL JOIN
(SELECT name, n as s3_n, 3 as s3_2 FROM t3) s3;

SELECT * FROM
(SELECT name, n as s1_n, 1 as s1_1 FROM t1) as s1
NATURAL INNER JOIN
(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2
NATURAL INNER JOIN
(SELECT name, n as s3_n, 3 as s3_2 FROM t3) s3;

SELECT * FROM
(SELECT name, n as s1_n, 1 as s1_1 FROM t1) as s1
NATURAL FULL JOIN
(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2
NATURAL FULL JOIN
(SELECT name, n as s3_n, 3 as s3_2 FROM t3) s3;

SELECT * FROM
(SELECT name, n as s1_n FROM t1) as s1
NATURAL FULL JOIN
  (SELECT * FROM
    (SELECT name, n as s2_n FROM t2) as s2
    NATURAL FULL JOIN
    (SELECT name, n as s3_n FROM t3) as s3
  ) ss2;

SELECT * FROM
(SELECT name, n as s1_n FROM t1) as s1
NATURAL FULL JOIN
  (SELECT * FROM
    (SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2
    NATURAL FULL JOIN
    (SELECT name, n as s3_n FROM t3) as s3
  ) ss2;


-- Test for propagation of nullability constraints into sub-joins

create temp table x (x1 int, x2 int);
insert into x values (1,11);
insert into x values (2,22);
insert into x values (3,null);
insert into x values (4,44);
insert into x values (5,null);

create temp table y (y1 int, y2 int);
insert into y values (1,111);
insert into y values (2,222);
insert into y values (3,333);
insert into y values (4,null);

select * from x;
select * from y;

select * from x left join y on (x1 = y1 and x2 is not null);
select * from x left join y on (x1 = y1 and y2 is not null);

select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1);
select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1 and x2 is not null);
select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1 and y2 is not null);
select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1 and xx2 is not null);
-- these should NOT give the same answers as above
select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1) where (x2 is not null);
select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1) where (y2 is not null);
select * from (x left join y on (x1 = y1)) left join x xx(xx1,xx2)
on (x1 = xx1) where (xx2 is not null);

--
-- regression test: check for bug with propagation of implied equality
-- to outside an IN
--
select count(*) from tenk1 a where unique1 in
  (select unique1 from tenk1 b join tenk1 c using (unique1)
   where b.unique2 = 42);

--
-- regression test: check for failure to generate a plan with multiple
-- degenerate IN clauses
--
select count(*) from tenk1 x where
  x.unique1 in (select a.f1 from int4_tbl a,float8_tbl b where a.f1=b.f1) and
  x.unique1 = 0 and
  x.unique1 in (select aa.f1 from int4_tbl aa,float8_tbl bb where aa.f1=bb.f1);

-- try that with GEQO too
begin;
set geqo = on;
set geqo_threshold = 2;
select count(*) from tenk1 x where
  x.unique1 in (select a.f1 from int4_tbl a,float8_tbl b where a.f1=b.f1) and
  x.unique1 = 0 and
  x.unique1 in (select aa.f1 from int4_tbl aa,float8_tbl bb where aa.f1=bb.f1);
rollback;


--
-- Clean up
--

DROP TABLE t1;
DROP TABLE t2;
DROP TABLE t3;

DROP TABLE J1_TBL;
DROP TABLE J2_TBL;

-- Both DELETE and UPDATE allow the specification of additional tables
-- to "join" against to determine which rows should be modified.

CREATE TEMP TABLE t1 (a int, b int);
CREATE TEMP TABLE t2 (a int, b int);
CREATE TEMP TABLE t3 (x int, y int);

INSERT INTO t1 VALUES (5, 10);
INSERT INTO t1 VALUES (15, 20);
INSERT INTO t1 VALUES (100, 100);
INSERT INTO t1 VALUES (200, 1000);
INSERT INTO t2 VALUES (200, 2000);
INSERT INTO t3 VALUES (5, 20);
INSERT INTO t3 VALUES (6, 7);
INSERT INTO t3 VALUES (7, 8);
INSERT INTO t3 VALUES (500, 100);

DELETE FROM t3 USING t1 table1 WHERE t3.x = table1.a;
SELECT * FROM t3;
DELETE FROM t3 USING t1 JOIN t2 USING (a) WHERE t3.x > t1.a;
SELECT * FROM t3;
DELETE FROM t3 USING t3 t3_other WHERE t3.x = t3_other.x AND t3.y = t3_other.y;
SELECT * FROM t3;

-- Test join against inheritance tree

create temp table t2a () inherits (t2);

insert into t2a values (200, 2001);

select * from t1 left join t2 on (t1.a = t2.a);

--
-- regression test for 8.1 merge right join bug
--

CREATE TEMP TABLE tt1 ( tt1_id int4, joincol int4 );
INSERT INTO tt1 VALUES (1, 11);
INSERT INTO tt1 VALUES (2, NULL);

CREATE TEMP TABLE tt2 ( tt2_id int4, joincol int4 );
INSERT INTO tt2 VALUES (21, 11);
INSERT INTO tt2 VALUES (22, 11);

set enable_hashjoin to off;
set enable_nestloop to off;

-- these should give the same results

select tt1.*, tt2.* from tt1 left join tt2 on tt1.joincol = tt2.joincol;

select tt1.*, tt2.* from tt2 right join tt1 on tt1.joincol = tt2.joincol;

reset enable_hashjoin;
reset enable_nestloop;

--
-- regression test for 8.2 bug with improper re-ordering of left joins
--

create temp table tt3(f1 int, f2 text);
insert into tt3 select x, repeat('xyzzy', 100) from generate_series(1,10000) x;
create index tt3i on tt3(f1);
analyze tt3;

create temp table tt4(f1 int);
insert into tt4 values (0),(1),(9999);
analyze tt4;

SELECT a.f1
FROM tt4 a
LEFT JOIN (
        SELECT b.f1
        FROM tt3 b LEFT JOIN tt3 c ON (b.f1 = c.f1)
        WHERE c.f1 IS NULL
) AS d ON (a.f1 = d.f1)
WHERE d.f1 IS NULL;

--
-- regression test for problems of the sort depicted in bug #3494
--

create temp table tt5(f1 int, f2 int);
create temp table tt6(f1 int, f2 int);

insert into tt5 values(1, 10);
insert into tt5 values(1, 11);

insert into tt6 values(1, 9);
insert into tt6 values(1, 2);
insert into tt6 values(2, 9);

select * from tt5,tt6 where tt5.f1 = tt6.f1 and tt5.f1 = tt5.f2 - tt6.f2;

--
-- regression test for problems of the sort depicted in bug #3588
--

create temp table xx (pkxx int);
create temp table yy (pkyy int, pkxx int);

insert into xx values (1);
insert into xx values (2);
insert into xx values (3);

insert into yy values (101, 1);
insert into yy values (201, 2);
insert into yy values (301, NULL);

select yy.pkyy as yy_pkyy, yy.pkxx as yy_pkxx, yya.pkyy as yya_pkyy,
       xxa.pkxx as xxa_pkxx, xxb.pkxx as xxb_pkxx
from yy
     left join (SELECT * FROM yy where pkyy = 101) as yya ON yy.pkyy = yya.pkyy
     left join xx xxa on yya.pkxx = xxa.pkxx
     left join xx xxb on coalesce (xxa.pkxx, 1) = xxb.pkxx;

--
-- regression test for improper pushing of constants across outer-join clauses
-- (as seen in early 8.2.x releases)
--

create temp table zt1 (f1 int primary key);
create temp table zt2 (f2 int primary key);
create temp table zt3 (f3 int primary key);
insert into zt1 values(53);
insert into zt2 values(53);

select * from
  zt2 left join zt3 on (f2 = f3)
      left join zt1 on (f3 = f1)
where f2 = 53;

create temp view zv1 as select *,'dummy'::text AS junk from zt1;

select * from
  zt2 left join zt3 on (f2 = f3)
      left join zv1 on (f3 = f1)
where f2 = 53;

--
-- regression test for improper extraction of OR indexqual conditions
-- (as seen in early 8.3.x releases)
--

select a.unique2, a.ten, b.tenthous, b.unique2, b.hundred
from tenk1 a left join tenk1 b on a.unique2 = b.tenthous
where a.unique1 = 42 and
      ((b.unique2 is null and a.ten = 2) or b.hundred = 3);

--
-- test proper positioning of one-time quals in EXISTS (8.4devel bug)
--
prepare foo(bool) as
  select count(*) from tenk1 a left join tenk1 b
    on (a.unique2 = b.unique1 and exists
        (select 1 from tenk1 c where c.thousand = b.unique2 and $1));
execute foo(true);
execute foo(false);

--
-- test for sane behavior with noncanonical merge clauses, per bug #4926
--

begin;

set enable_mergejoin = 1;
set enable_hashjoin = 0;
set enable_nestloop = 0;

create temp table a (i integer);
create temp table b (x integer, y integer);

select * from a left join b on i = x and i = y and x = i;

rollback;

--
-- test NULL behavior of whole-row Vars, per bug #5025
--
select t1.q2, count(t2.*)
from int8_tbl t1 left join int8_tbl t2 on (t1.q2 = t2.q1)
group by t1.q2 order by 1;

select t1.q2, count(t2.*)
from int8_tbl t1 left join (select * from int8_tbl) t2 on (t1.q2 = t2.q1)
group by t1.q2 order by 1;

select t1.q2, count(t2.*)
from int8_tbl t1 left join (select * from int8_tbl offset 0) t2 on (t1.q2 = t2.q1)
group by t1.q2 order by 1;

select t1.q2, count(t2.*)
from int8_tbl t1 left join
  (select q1, case when q2=1 then 1 else q2 end as q2 from int8_tbl) t2
  on (t1.q2 = t2.q1)
group by t1.q2 order by 1;

--
-- test incorrect failure to NULL pulled-up subexpressions
--
begin;

create temp table a (
     code char not null,
     constraint a_pk primary key (code)
);
create temp table b (
     a char not null,
     num integer not null,
     constraint b_pk primary key (a, num)
);
create temp table c (
     name char not null,
     a char,
     constraint c_pk primary key (name)
);

insert into a (code) values ('p');
insert into a (code) values ('q');
insert into b (a, num) values ('p', 1);
insert into b (a, num) values ('p', 2);
insert into c (name, a) values ('A', 'p');
insert into c (name, a) values ('B', 'q');
insert into c (name, a) values ('C', null);

select c.name, ss.code, ss.b_cnt, ss.const
from c left join
  (select a.code, coalesce(b_grp.cnt, 0) as b_cnt, -1 as const
   from a left join
     (select count(1) as cnt, b.a from b group by b.a) as b_grp
     on a.code = b_grp.a
  ) as ss
  on (c.a = ss.code)
order by c.name;

rollback;

--
-- test incorrect handling of placeholders that only appear in targetlists,
-- per bug #6154
--
SELECT * FROM
( SELECT 1 as key1 ) sub1
LEFT JOIN
( SELECT sub3.key3, sub4.value2, COALESCE(sub4.value2, 66) as value3 FROM
    ( SELECT 1 as key3 ) sub3
    LEFT JOIN
    ( SELECT sub5.key5, COALESCE(sub6.value1, 1) as value2 FROM
        ( SELECT 1 as key5 ) sub5
        LEFT JOIN
        ( SELECT 2 as key6, 42 as value1 ) sub6
        ON sub5.key5 = sub6.key6
    ) sub4
    ON sub4.key5 = sub3.key3
) sub2
ON sub1.key1 = sub2.key3;

-- test the path using join aliases, too
SELECT * FROM
( SELECT 1 as key1 ) sub1
LEFT JOIN
( SELECT sub3.key3, value2, COALESCE(value2, 66) as value3 FROM
    ( SELECT 1 as key3 ) sub3
    LEFT JOIN
    ( SELECT sub5.key5, COALESCE(sub6.value1, 1) as value2 FROM
        ( SELECT 1 as key5 ) sub5
        LEFT JOIN
        ( SELECT 2 as key6, 42 as value1 ) sub6
        ON sub5.key5 = sub6.key6
    ) sub4
    ON sub4.key5 = sub3.key3
) sub2
ON sub1.key1 = sub2.key3;

--
-- test case where a PlaceHolderVar is used as a nestloop parameter
--

EXPLAIN (COSTS OFF)
SELECT qq, unique1
  FROM
  ( SELECT COALESCE(q1, 0) AS qq FROM int8_tbl a ) AS ss1
  FULL OUTER JOIN
  ( SELECT COALESCE(q2, -1) AS qq FROM int8_tbl b ) AS ss2
  USING (qq)
  INNER JOIN tenk1 c ON qq = unique2;

SELECT qq, unique1
  FROM
  ( SELECT COALESCE(q1, 0) AS qq FROM int8_tbl a ) AS ss1
  FULL OUTER JOIN
  ( SELECT COALESCE(q2, -1) AS qq FROM int8_tbl b ) AS ss2
  USING (qq)
  INNER JOIN tenk1 c ON qq = unique2;

--
-- nested nestloops can require nested PlaceHolderVars
--

create temp table nt1 (
  id int primary key,
  a1 boolean,
  a2 boolean
);
create temp table nt2 (
  id int primary key,
  nt1_id int,
  b1 boolean,
  b2 boolean,
  foreign key (nt1_id) references nt1(id)
);
create temp table nt3 (
  id int primary key,
  nt2_id int,
  c1 boolean,
  foreign key (nt2_id) references nt2(id)
);

insert into nt1 values (1,true,true);
insert into nt1 values (2,true,false);
insert into nt1 values (3,false,false);
insert into nt2 values (1,1,true,true);
insert into nt2 values (2,2,true,false);
insert into nt2 values (3,3,false,false);
insert into nt3 values (1,1,true);
insert into nt3 values (2,2,false);
insert into nt3 values (3,3,true);

explain (costs off)
select nt3.id
from nt3 as nt3
  left join
    (select nt2.*, (nt2.b1 and ss1.a3) AS b3
     from nt2 as nt2
       left join
         (select nt1.*, (nt1.id is not null) as a3 from nt1) as ss1
         on ss1.id = nt2.nt1_id
    ) as ss2
    on ss2.id = nt3.nt2_id
where nt3.id = 1 and ss2.b3;

select nt3.id
from nt3 as nt3
  left join
    (select nt2.*, (nt2.b1 and ss1.a3) AS b3
     from nt2 as nt2
       left join
         (select nt1.*, (nt1.id is not null) as a3 from nt1) as ss1
         on ss1.id = nt2.nt1_id
    ) as ss2
    on ss2.id = nt3.nt2_id
where nt3.id = 1 and ss2.b3;

--
-- test case where a PlaceHolderVar is propagated into a subquery
--

explain (costs off)
select * from
  int8_tbl t1 left join
  (select q1 as x, 42 as y from int8_tbl t2) ss
  on t1.q2 = ss.x
where
  1 = (select 1 from int8_tbl t3 where ss.y is not null limit 1)
order by 1,2;

select * from
  int8_tbl t1 left join
  (select q1 as x, 42 as y from int8_tbl t2) ss
  on t1.q2 = ss.x
where
  1 = (select 1 from int8_tbl t3 where ss.y is not null limit 1)
order by 1,2;

--
-- test the corner cases FULL JOIN ON TRUE and FULL JOIN ON FALSE
--
select * from int4_tbl a full join int4_tbl b on true;
select * from int4_tbl a full join int4_tbl b on false;

--
-- test for ability to use a cartesian join when necessary
--

explain (costs off)
select * from
  tenk1 join int4_tbl on f1 = twothousand,
  int4(sin(1)) q1,
  int4(sin(0)) q2
where q1 = thousand or q2 = thousand;

explain (costs off)
select * from
  tenk1 join int4_tbl on f1 = twothousand,
  int4(sin(1)) q1,
  int4(sin(0)) q2
where thousand = (q1 + q2);

--
-- test extraction of restriction OR clauses from join OR clause
-- (we used to only do this for indexable clauses)
--

explain (costs off)
select * from tenk1 a join tenk1 b on
  (a.unique1 = 1 and b.unique1 = 2) or (a.unique2 = 3 and b.hundred = 4);
explain (costs off)
select * from tenk1 a join tenk1 b on
  (a.unique1 = 1 and b.unique1 = 2) or (a.unique2 = 3 and b.ten = 4);
explain (costs off)
select * from tenk1 a join tenk1 b on
  (a.unique1 = 1 and b.unique1 = 2) or
  ((a.unique2 = 3 or a.unique2 = 7) and b.hundred = 4);

--
-- test placement of movable quals in a parameterized join tree
--

explain (costs off)
select * from tenk1 t1 left join
  (tenk1 t2 join tenk1 t3 on t2.thousand = t3.unique2)
  on t1.hundred = t2.hundred and t1.ten = t3.ten
where t1.unique1 = 1;

explain (costs off)
select * from tenk1 t1 left join
  (tenk1 t2 join tenk1 t3 on t2.thousand = t3.unique2)
  on t1.hundred = t2.hundred and t1.ten + t2.ten = t3.ten
where t1.unique1 = 1;

explain (costs off)
select count(*) from
  tenk1 a join tenk1 b on a.unique1 = b.unique2
  left join tenk1 c on a.unique2 = b.unique1 and c.thousand = a.thousand
  join int4_tbl on b.thousand = f1;

select count(*) from
  tenk1 a join tenk1 b on a.unique1 = b.unique2
  left join tenk1 c on a.unique2 = b.unique1 and c.thousand = a.thousand
  join int4_tbl on b.thousand = f1;

explain (costs off)
select b.unique1 from
  tenk1 a join tenk1 b on a.unique1 = b.unique2
  left join tenk1 c on b.unique1 = 42 and c.thousand = a.thousand
  join int4_tbl i1 on b.thousand = f1
  right join int4_tbl i2 on i2.f1 = b.tenthous
  order by 1;

select b.unique1 from
  tenk1 a join tenk1 b on a.unique1 = b.unique2
  left join tenk1 c on b.unique1 = 42 and c.thousand = a.thousand
  join int4_tbl i1 on b.thousand = f1
  right join int4_tbl i2 on i2.f1 = b.tenthous
  order by 1;

explain (costs off)
select * from
(
  select unique1, q1, coalesce(unique1, -1) + q1 as fault
  from int8_tbl left join tenk1 on (q2 = unique2)
) ss
where fault = 122
order by fault;

select * from
(
  select unique1, q1, coalesce(unique1, -1) + q1 as fault
  from int8_tbl left join tenk1 on (q2 = unique2)
) ss
where fault = 122
order by fault;

--
-- test handling of potential equivalence clauses above outer joins
--

explain (costs off)
select q1, unique2, thousand, hundred
  from int8_tbl a left join tenk1 b on q1 = unique2
  where coalesce(thousand,123) = q1 and q1 = coalesce(hundred,123);

select q1, unique2, thousand, hundred
  from int8_tbl a left join tenk1 b on q1 = unique2
  where coalesce(thousand,123) = q1 and q1 = coalesce(hundred,123);

explain (costs off)
select f1, unique2, case when unique2 is null then f1 else 0 end
  from int4_tbl a left join tenk1 b on f1 = unique2
  where (case when unique2 is null then f1 else 0 end) = 0;

select f1, unique2, case when unique2 is null then f1 else 0 end
  from int4_tbl a left join tenk1 b on f1 = unique2
  where (case when unique2 is null then f1 else 0 end) = 0;

--
-- another case with equivalence clauses above outer joins (bug #8591)
--

explain (costs off)
select a.unique1, b.unique1, c.unique1, coalesce(b.twothousand, a.twothousand)
  from tenk1 a left join tenk1 b on b.thousand = a.unique1                        left join tenk1 c on c.unique2 = coalesce(b.twothousand, a.twothousand)
  where a.unique2 = 5530 and coalesce(b.twothousand, a.twothousand) = 44;

select a.unique1, b.unique1, c.unique1, coalesce(b.twothousand, a.twothousand)
  from tenk1 a left join tenk1 b on b.thousand = a.unique1                        left join tenk1 c on c.unique2 = coalesce(b.twothousand, a.twothousand)
  where a.unique2 = 5530 and coalesce(b.twothousand, a.twothousand) = 44;

--
-- check handling of join aliases when flattening multiple levels of subquery
--

explain (verbose, costs off)
select foo1.join_key as foo1_id, foo3.join_key AS foo3_id, bug_field from
  (values (0),(1)) foo1(join_key)
left join
  (select join_key, bug_field from
    (select ss1.join_key, ss1.bug_field from
      (select f1 as join_key, 666 as bug_field from int4_tbl i1) ss1
    ) foo2
   left join
    (select unique2 as join_key from tenk1 i2) ss2
   using (join_key)
  ) foo3
using (join_key);

select foo1.join_key as foo1_id, foo3.join_key AS foo3_id, bug_field from
  (values (0),(1)) foo1(join_key)
left join
  (select join_key, bug_field from
    (select ss1.join_key, ss1.bug_field from
      (select f1 as join_key, 666 as bug_field from int4_tbl i1) ss1
    ) foo2
   left join
    (select unique2 as join_key from tenk1 i2) ss2
   using (join_key)
  ) foo3
using (join_key);

--
-- test ability to push constants through outer join clauses
--

explain (costs off)
  select * from int4_tbl a left join tenk1 b on f1 = unique2 where f1 = 0;

explain (costs off)
  select * from tenk1 a full join tenk1 b using(unique2) where unique2 = 42;

--
-- test join removal
--

begin;

CREATE TEMP TABLE a (id int PRIMARY KEY, b_id int);
CREATE TEMP TABLE b (id int PRIMARY KEY, c_id int);
CREATE TEMP TABLE c (id int PRIMARY KEY);
CREATE TEMP TABLE d (a int, b int);
INSERT INTO a VALUES (0, 0), (1, NULL);
INSERT INTO b VALUES (0, 0), (1, NULL);
INSERT INTO c VALUES (0), (1);
INSERT INTO d VALUES (1,3), (2,2), (3,1);

-- all three cases should be optimizable into a simple seqscan
explain (costs off) SELECT a.* FROM a LEFT JOIN b ON a.b_id = b.id;
explain (costs off) SELECT b.* FROM b LEFT JOIN c ON b.c_id = c.id;
explain (costs off)
  SELECT a.* FROM a LEFT JOIN (b left join c on b.c_id = c.id)
  ON (a.b_id = b.id);

-- check optimization of outer join within another special join
explain (costs off)
select id from a where id in (
	select b.id from b left join c on b.id = c.id
);

-- check that join removal works for a left join when joining a subquery
-- that is guaranteed to be unique by its GROUP BY clause
explain (costs off)
select d.* from d left join (select * from b group by b.id, b.c_id) s
  on d.a = s.id and d.b = s.c_id;

-- similarly, but keying off a DISTINCT clause
explain (costs off)
select d.* from d left join (select distinct * from b) s
  on d.a = s.id and d.b = s.c_id;

-- join removal is not possible when the GROUP BY contains a column that is
-- not in the join condition
explain (costs off)
select d.* from d left join (select * from b group by b.id, b.c_id) s
  on d.a = s.id;

-- similarly, but keying off a DISTINCT clause
explain (costs off)
select d.* from d left join (select distinct * from b) s
  on d.a = s.id;

-- check join removal works when uniqueness of the join condition is enforced
-- by a UNION
explain (costs off)
select d.* from d left join (select id from a union select id from b) s
  on d.a = s.id;

-- check join removal with a cross-type comparison operator
explain (costs off)
select i8.* from int8_tbl i8 left join (select f1 from int4_tbl group by f1) i4
  on i8.q1 = i4.f1;

rollback;

create temp table parent (k int primary key, pd int);
create temp table child (k int unique, cd int);
insert into parent values (1, 10), (2, 20), (3, 30);
insert into child values (1, 100), (4, 400);

-- this case is optimizable
select p.* from parent p left join child c on (p.k = c.k);
explain (costs off)
  select p.* from parent p left join child c on (p.k = c.k);

-- this case is not
select p.*, linked from parent p
  left join (select c.*, true as linked from child c) as ss
  on (p.k = ss.k);
explain (costs off)
  select p.*, linked from parent p
    left join (select c.*, true as linked from child c) as ss
    on (p.k = ss.k);

-- check for a 9.0rc1 bug: join removal breaks pseudoconstant qual handling
select p.* from
  parent p left join child c on (p.k = c.k)
  where p.k = 1 and p.k = 2;
explain (costs off)
select p.* from
  parent p left join child c on (p.k = c.k)
  where p.k = 1 and p.k = 2;

select p.* from
  (parent p left join child c on (p.k = c.k)) join parent x on p.k = x.k
  where p.k = 1 and p.k = 2;
explain (costs off)
select p.* from
  (parent p left join child c on (p.k = c.k)) join parent x on p.k = x.k
  where p.k = 1 and p.k = 2;

-- bug 5255: this is not optimizable by join removal
begin;

CREATE TEMP TABLE a (id int PRIMARY KEY);
CREATE TEMP TABLE b (id int PRIMARY KEY, a_id int);
INSERT INTO a VALUES (0), (1);
INSERT INTO b VALUES (0, 0), (1, NULL);

SELECT * FROM b LEFT JOIN a ON (b.a_id = a.id) WHERE (a.id IS NULL OR a.id > 0);
SELECT b.* FROM b LEFT JOIN a ON (b.a_id = a.id) WHERE (a.id IS NULL OR a.id > 0);

rollback;

-- another join removal bug: this is not optimizable, either
begin;

create temp table innertab (id int8 primary key, dat1 int8);
insert into innertab values(123, 42);

SELECT * FROM
    (SELECT 1 AS x) ss1
  LEFT JOIN
    (SELECT q1, q2, COALESCE(dat1, q1) AS y
     FROM int8_tbl LEFT JOIN innertab ON q2 = id) ss2
  ON true;

rollback;

-- bug #8444: we've historically allowed duplicate aliases within aliased JOINs

select * from
  int8_tbl x join (int4_tbl x cross join int4_tbl y) j on q1 = f1; -- error
select * from
  int8_tbl x join (int4_tbl x cross join int4_tbl y) j on q1 = y.f1; -- error
select * from
  int8_tbl x join (int4_tbl x cross join int4_tbl y(ff)) j on q1 = f1; -- ok

--
-- Test LATERAL
--

select unique2, x.*
from tenk1 a, lateral (select * from int4_tbl b where f1 = a.unique1) x;
explain (costs off)
  select unique2, x.*
  from tenk1 a, lateral (select * from int4_tbl b where f1 = a.unique1) x;
select unique2, x.*
from int4_tbl x, lateral (select unique2 from tenk1 where f1 = unique1) ss;
explain (costs off)
  select unique2, x.*
  from int4_tbl x, lateral (select unique2 from tenk1 where f1 = unique1) ss;
explain (costs off)
  select unique2, x.*
  from int4_tbl x cross join lateral (select unique2 from tenk1 where f1 = unique1) ss;
select unique2, x.*
from int4_tbl x left join lateral (select unique1, unique2 from tenk1 where f1 = unique1) ss on true;
explain (costs off)
  select unique2, x.*
  from int4_tbl x left join lateral (select unique1, unique2 from tenk1 where f1 = unique1) ss on true;

-- check scoping of lateral versus parent references
-- the first of these should return int8_tbl.q2, the second int8_tbl.q1
select *, (select r from (select q1 as q2) x, (select q2 as r) y) from int8_tbl;
select *, (select r from (select q1 as q2) x, lateral (select q2 as r) y) from int8_tbl;

-- lateral with function in FROM
select count(*) from tenk1 a, lateral generate_series(1,two) g;
explain (costs off)
  select count(*) from tenk1 a, lateral generate_series(1,two) g;
explain (costs off)
  select count(*) from tenk1 a cross join lateral generate_series(1,two) g;
-- don't need the explicit LATERAL keyword for functions
explain (costs off)
  select count(*) from tenk1 a, generate_series(1,two) g;

-- lateral with UNION ALL subselect
explain (costs off)
  select * from generate_series(100,200) g,
    lateral (select * from int8_tbl a where g = q1 union all
             select * from int8_tbl b where g = q2) ss;
select * from generate_series(100,200) g,
  lateral (select * from int8_tbl a where g = q1 union all
           select * from int8_tbl b where g = q2) ss;

-- lateral with VALUES
explain (costs off)
  select count(*) from tenk1 a,
    tenk1 b join lateral (values(a.unique1)) ss(x) on b.unique2 = ss.x;
select count(*) from tenk1 a,
  tenk1 b join lateral (values(a.unique1)) ss(x) on b.unique2 = ss.x;

-- lateral injecting a strange outer join condition
explain (costs off)
  select * from int8_tbl a,
    int8_tbl x left join lateral (select a.q1 from int4_tbl y) ss(z)
      on x.q2 = ss.z;
select * from int8_tbl a,
  int8_tbl x left join lateral (select a.q1 from int4_tbl y) ss(z)
    on x.q2 = ss.z;

-- lateral reference to a join alias variable
select * from (select f1/2 as x from int4_tbl) ss1 join int4_tbl i4 on x = f1,
  lateral (select x) ss2(y);
select * from (select f1 as x from int4_tbl) ss1 join int4_tbl i4 on x = f1,
  lateral (values(x)) ss2(y);
select * from ((select f1/2 as x from int4_tbl) ss1 join int4_tbl i4 on x = f1) j,
  lateral (select x) ss2(y);

-- lateral references requiring pullup
select * from (values(1)) x(lb),
  lateral generate_series(lb,4) x4;
select * from (select f1/1000000000 from int4_tbl) x(lb),
  lateral generate_series(lb,4) x4;
select * from (values(1)) x(lb),
  lateral (values(lb)) y(lbcopy);
select * from (values(1)) x(lb),
  lateral (select lb from int4_tbl) y(lbcopy);
select * from
  int8_tbl x left join (select q1,coalesce(q2,0) q2 from int8_tbl) y on x.q2 = y.q1,
  lateral (values(x.q1,y.q1,y.q2)) v(xq1,yq1,yq2);
select * from
  int8_tbl x left join (select q1,coalesce(q2,0) q2 from int8_tbl) y on x.q2 = y.q1,
  lateral (select x.q1,y.q1,y.q2) v(xq1,yq1,yq2);
select x.* from
  int8_tbl x left join (select q1,coalesce(q2,0) q2 from int8_tbl) y on x.q2 = y.q1,
  lateral (select x.q1,y.q1,y.q2) v(xq1,yq1,yq2);
select v.* from
  (int8_tbl x left join (select q1,coalesce(q2,0) q2 from int8_tbl) y on x.q2 = y.q1)
  left join int4_tbl z on z.f1 = x.q2,
  lateral (select x.q1,y.q1 union all select x.q2,y.q2) v(vx,vy);
select v.* from
  (int8_tbl x left join (select q1,(select coalesce(q2,0)) q2 from int8_tbl) y on x.q2 = y.q1)
  left join int4_tbl z on z.f1 = x.q2,
  lateral (select x.q1,y.q1 union all select x.q2,y.q2) v(vx,vy);
create temp table dual();
insert into dual default values;
analyze dual;
select v.* from
  (int8_tbl x left join (select q1,(select coalesce(q2,0)) q2 from int8_tbl) y on x.q2 = y.q1)
  left join int4_tbl z on z.f1 = x.q2,
  lateral (select x.q1,y.q1 from dual union all select x.q2,y.q2 from dual) v(vx,vy);

explain (verbose, costs off)
select * from
  int8_tbl a left join
  lateral (select *, a.q2 as x from int8_tbl b) ss on a.q2 = ss.q1;
select * from
  int8_tbl a left join
  lateral (select *, a.q2 as x from int8_tbl b) ss on a.q2 = ss.q1;
explain (verbose, costs off)
select * from
  int8_tbl a left join
  lateral (select *, coalesce(a.q2, 42) as x from int8_tbl b) ss on a.q2 = ss.q1;
select * from
  int8_tbl a left join
  lateral (select *, coalesce(a.q2, 42) as x from int8_tbl b) ss on a.q2 = ss.q1;

-- lateral can result in join conditions appearing below their
-- real semantic level
explain (verbose, costs off)
select * from int4_tbl i left join
  lateral (select * from int2_tbl j where i.f1 = j.f1) k on true;
select * from int4_tbl i left join
  lateral (select * from int2_tbl j where i.f1 = j.f1) k on true;
explain (verbose, costs off)
select * from int4_tbl i left join
  lateral (select coalesce(i) from int2_tbl j where i.f1 = j.f1) k on true;
select * from int4_tbl i left join
  lateral (select coalesce(i) from int2_tbl j where i.f1 = j.f1) k on true;
explain (verbose, costs off)
select * from int4_tbl a,
  lateral (
    select * from int4_tbl b left join int8_tbl c on (b.f1 = q1 and a.f1 = q2)
  ) ss;
select * from int4_tbl a,
  lateral (
    select * from int4_tbl b left join int8_tbl c on (b.f1 = q1 and a.f1 = q2)
  ) ss;

-- lateral reference in a PlaceHolderVar evaluated at join level
explain (verbose, costs off)
select * from
  int8_tbl a left join lateral
  (select b.q1 as bq1, c.q1 as cq1, least(a.q1,b.q1,c.q1) from
   int8_tbl b cross join int8_tbl c) ss
  on a.q2 = ss.bq1;
select * from
  int8_tbl a left join lateral
  (select b.q1 as bq1, c.q1 as cq1, least(a.q1,b.q1,c.q1) from
   int8_tbl b cross join int8_tbl c) ss
  on a.q2 = ss.bq1;

-- case requiring nested PlaceHolderVars
explain (verbose, costs off)
select * from
  int8_tbl c left join (
    int8_tbl a left join (select q1, coalesce(q2,42) as x from int8_tbl b) ss1
      on a.q2 = ss1.q1
    cross join
    lateral (select q1, coalesce(ss1.x,q2) as y from int8_tbl d) ss2
  ) on c.q2 = ss2.q1,
  lateral (select ss2.y) ss3;

-- case that breaks the old ph_may_need optimization
explain (verbose, costs off)
select c.*,a.*,ss1.q1,ss2.q1,ss3.* from
  int8_tbl c left join (
    int8_tbl a left join
      (select q1, coalesce(q2,f1) as x from int8_tbl b, int4_tbl b2
       where q1 < f1) ss1
      on a.q2 = ss1.q1
    cross join
    lateral (select q1, coalesce(ss1.x,q2) as y from int8_tbl d) ss2
  ) on c.q2 = ss2.q1,
  lateral (select * from int4_tbl i where ss2.y > f1) ss3;

-- check processing of postponed quals (bug #9041)
explain (verbose, costs off)
select * from
  (select 1 as x) x cross join (select 2 as y) y
  left join lateral (
    select * from (select 3 as z) z where z.z = x.x
  ) zz on zz.z = y.y;

-- test some error cases where LATERAL should have been used but wasn't
select f1,g from int4_tbl a, (select f1 as g) ss;
select f1,g from int4_tbl a, (select a.f1 as g) ss;
select f1,g from int4_tbl a cross join (select f1 as g) ss;
select f1,g from int4_tbl a cross join (select a.f1 as g) ss;
-- SQL:2008 says the left table is in scope but illegal to access here
select f1,g from int4_tbl a right join lateral generate_series(0, a.f1) g on true;
select f1,g from int4_tbl a full join lateral generate_series(0, a.f1) g on true;
-- check we complain about ambiguous table references
select * from
  int8_tbl x cross join (int4_tbl x cross join lateral (select x.f1) ss);
-- LATERAL can be used to put an aggregate into the FROM clause of its query
select 1 from tenk1 a, lateral (select max(a.unique1) from int4_tbl b) ss;

-- check behavior of LATERAL in UPDATE/DELETE

create temp table xx1 as select f1 as x1, -f1 as x2 from int4_tbl;

-- error, can't do this:
update xx1 set x2 = f1 from (select * from int4_tbl where f1 = x1) ss;
update xx1 set x2 = f1 from (select * from int4_tbl where f1 = xx1.x1) ss;
-- can't do it even with LATERAL:
update xx1 set x2 = f1 from lateral (select * from int4_tbl where f1 = x1) ss;
-- we might in future allow something like this, but for now it's an error:
update xx1 set x2 = f1 from xx1, lateral (select * from int4_tbl where f1 = x1) ss;

-- also errors:
delete from xx1 using (select * from int4_tbl where f1 = x1) ss;
delete from xx1 using (select * from int4_tbl where f1 = xx1.x1) ss;
delete from xx1 using lateral (select * from int4_tbl where f1 = x1) ss;
