text: --
  -- 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;
  
  --
  -- 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);
  
  --
  -- test placement of movable quals in a parameterized join tree
  --
  
  set effective_cache_size = '128MB';
  
  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);
  INSERT INTO a VALUES (0, 0), (1, NULL);
  INSERT INTO b VALUES (0, 0), (1, NULL);
  INSERT INTO c VALUES (0), (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
  );
  
  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;
  
  -- 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;
  