--
-- 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 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 join removal
--

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);