我有两张桌子,foo并且bar:
+----+-----+ +----+-----+
| id | val | | id | val |
+----+-----+ +----+-----+
| 1 | qwe | | 1 | asd |
| 2 | rty | | 3 | fgh |
+----+-----+ +----+-----+
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id在这里不是唯一的.并非所有ID foo都具有其等价物,bar反之亦然.我需要计算两个表中具有特定ID的所有行,并将它们显示在新表中,例如:
+----+-----------+-----------+
| id | count_foo | count_bar |
+----+-----------+-----------+
| 1 | 1 | 1 |
| 2 | 1 | 0 |
| 3 | 0 | 1 |
+----+-----------+-----------+
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我试过了UNION SELECT:
SELECT id, COUNT(id) AS count_foo, 0 AS count_bar FROM foo GROUP BY id
UNION SELECT id, 0, COUNT(id) FROM bar GROUP BY id;
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但这会输出id=1两次,如
+----+-----------+-----------+
| id | count_foo | count_bar |
+----+-----------+-----------+
| 1 | 1 | 0 | <- not good
| 2 | 1 | 0 |
| 1 | 0 | 1 | <- not good
| 3 | 0 | 1 |
+----+-----------+-----------+
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我也尝试过LEFT JOIN:
SELECT id, COUNT(foo.id) AS count_foo, COUNT(bar.id) AS count_bar
FROM foo LEFT JOIN bar USING(id) GROUP BY id;
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但是此查询忽略表bar中缺少ID的表中的行foo:
+----+-----------+-----------+
| id | count_foo | count_bar |
+----+-----------+-----------+
| 1 | 1 | 1 |
| 2 | 1 | 0 |
+----+-----------+-----------+ <- can I haz `id=3`?
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我错过了什么?什么是正确的查询或正确的手册阅读?
谢谢.
您可能想尝试以下方法:
SELECT d.id,
(SELECT COUNT(*) FROM foo WHERE id = d.id) count_foo,
(SELECT COUNT(*) FROM bar WHERE id = d.id) count_bar
FROM ((SELECT id FROM foo) UNION (SELECT id FROM bar)) d;
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测试用例:
CREATE TABLE foo (id int, val varchar(5));
CREATE TABLE bar (id int, val varchar(5));
INSERT INTO foo VALUES (1, 'qwe');
INSERT INTO foo VALUES (2, 'rty');
INSERT INTO bar VALUES (1, 'asf');
INSERT INTO bar VALUES (3, 'ghj');
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结果:
+------+-----------+-----------+
| id | count_foo | count_bar |
+------+-----------+-----------+
| 1 | 1 | 1 |
| 2 | 1 | 0 |
| 3 | 0 | 1 |
+------+-----------+-----------+
3 rows in set (0.00 sec)
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