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组合两个select查询

left-join subquery join mysql

1

Brook Julias · 技术社区 · 15 年前

+--------------+    +--------------+    +--------------------------+
|    table_1   |    |    table_2   |    |          table_3         |
+----+---------+    +----+---------+    +----+----------+----------+
| id |   name  |    | id |   name  |    | id |  tbl1_id |  tbl2_id | 
+----+---------+    +----+---------+    +----+----------+----------+
| 1  | tbl1_1  |    | 1  | tbl2_1  |    | id |     1    |     1    |
| 2  | tbl1_2  |    | 2  | tbl2_2  |    | id |     3    |     2    |
| 3  | tbl1_3  |    | 3  | tbl2_3  |    | id |     3    |     3    |
| 4  | tbl1_4  |    +----+---------+    +----+----------+----------+
+----+---------+

两者之间存在着多对多的关系 table_1 和 table_2 table_3 . 到目前为止,我一直使用分隔查询。一个查询返回 表1 连接到 表1 通过 表3 JOIN :

SELECT table_1.id, table_1.name, table_2.id, table_2.name
FROM table_3
LEFT JOIN table_1 ON (table_3.tbl1_id = table_1.id)
LEFT JOIN table_1 ON (table_2.tbl2_id = table_2.id)

它返回了我想要的结果,只返回了忽略了 表1 . 我尝试过使用子查询:

SELECT  table_1.id,
    table_1.name,
    (SELECT table_2.id FROM table_2, table_3 WHERE table_2.id = table_3.tbl2_id AND table_1.id = table_3.tbl1_id) AS tbl_2_id,
    (SELECT table_2.name FROM table_2, table_3 WHERE table_2.id = table_3.tbl2_id AND table_1.id = table_3.tbl1_id) AS tbl_2_name
FROM table_1

ERROR 1242 . 到目前为止,我还没有得到任何工作。我想要的结果与此类似。

+---------------+---------------+---------------+---------------+
|table_1.id |table_1.name   |table_2.id |table_2.name   |
+---------------+---------------+---------------+---------------+
|      1    |    tbl1_1 |      1    |    tbl2_1 |
|      2    |    tbl1_2 |       |       |
|      3    |    tbl1_3 |      2    |    tbl2_2 |
|      3    |    tbl1_3 |      3    |    tbl2_3 |
|      4    |    tbl1_4 |       |       |
+---------------+---------------+---------------+---------------+

table_1.name 和 table_2.name . 如果有人有什么建议,请告诉我。

3 回复 | 直到 15 年前

1

Mark Byers 15 年前

要从表\u 1中获取在其他表中不匹配的行,应使用外部联接而不是内部联接:

SELECT
    table_1.id,
    table_1.name,
    table_2.id,
    table_2.name
FROM table_1
LEFT JOIN table_3 ON table_3.tbl1_id = table_1.id
LEFT JOIN table_2 ON table_3.tbl2_id = table_2.id

结果:

table_1.id  table_1.name  table_2.id  table_2.name
1           'tbl1_1'      1           'tbl2_1'    
2           'tbl1_2'                  ''          
3           'tbl1_3'      2           'tbl2_2'    
3           'tbl1_3'      3           'tbl2_3'    
4           'tbl1_4'                  ''

2

0

ngroot 15 年前

SELECT table_1.id, table_1.name, table_2.id, table_2.name
FROM table_3
INNER JOIN table_1 ON table_1.id = table_3.tbl1_id
INNER JOIN table_2 ON table_2.id = table_3.tbl2_id
ORDER BY table_1.id, table_2.id

3

0

Brook Julias 14 年前

LEFT JOIN 到 RIGHT JOIN 两个连接在查询中出现的顺序解决了这个问题。这是工作守则的副本;

SELECT  table_1.id,
    table_1.name,
    table_2.id,
    table_2.name
FROM    table_3
RIGHT JOIN  table_2 ON (table_3.tbl2_id = table_2.id)
RIGHT JOIN  table_1 ON (table_3.tbl1_id = table_1.id)
ORDER BY table_1.name ASC, table_2.name ASC;