join.sql
来自「postgresql8.3.4源码,开源数据库」· SQL 代码 · 共 498 行
SQL
498 行
---- 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 columnSELECT '' AS "xxx", i, k, t FROM J1_TBL CROSS JOIN J2_TBL;-- resolve previous ambiguity by specifying the table nameSELECT '' 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 columnSELECT '' AS "xxx", * FROM J1_TBL INNER JOIN J2_TBL USING (i);-- Same as above, slightly different syntaxSELECT '' 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 namesSELECT '' 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 ( 'aa', 11 );INSERT INTO t2 VALUES ( 'aa', 12 );INSERT INTO t2 VALUES ( 'bb', 22 );INSERT INTO t2 VALUES ( 'dd', 42 );INSERT INTO t3 VALUES ( 'aa', 13 );INSERT INTO t3 VALUES ( 'bb', 23 );INSERT INTO t3 VALUES ( 'cc', 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 s2INNER JOIN(SELECT * FROM t3) s3USING (name);SELECT * FROM(SELECT * FROM t2) as s2LEFT JOIN(SELECT * FROM t3) s3USING (name);SELECT * FROM(SELECT * FROM t2) as s2FULL JOIN(SELECT * FROM t3) s3USING (name);-- Cases with non-nullable expressions in subquery results;-- make sure these go to null as expectedSELECT * FROM(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2NATURAL 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 s2NATURAL 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 s2NATURAL 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 s1NATURAL INNER JOIN(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2NATURAL 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 s1NATURAL FULL JOIN(SELECT name, n as s2_n, 2 as s2_2 FROM t2) as s2NATURAL 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 s1NATURAL 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 s1NATURAL 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-joinscreate 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 aboveselect * 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);---- 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;---- 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 resultsselect 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.f1FROM tt4 aLEFT 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_pkxxfrom 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.hundredfrom tenk1 a left join tenk1 b on a.unique2 = b.tenthouswhere a.unique1 = 42 and ((b.unique2 is null and a.ten = 2) or b.hundred = 3);
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