Question

我有一个（遗留）表，其中包含列：

bug_num   build_id    closed_to
1         3            NULL
2         4            NULL
3         NULL         1
4         3            NULL
5         NULL         2

我想编写一个查询，它将从特定版本中选择所有错误，以及所有对该版本中的错误关闭的错误。因此，如果我想为构建3执行此操作，它将包括#s 1和4（因为它们位于构建3中）以及3，因为它关闭了构建3（1）中的错误。

我以为我很接近：

SELECT stat.bug_num, 
       stat.build_id 
FROM   bug_status stat 
       JOIN bug_status stat2 
         ON stat2.closed_to = stat.bug_num 
WHERE  stat.build_id = 3;

......但它似乎没有给我想要的结果。谢谢你的帮助！

Answer 1

您没有在WHERE子句中包含stat2.build_id（我认为您的ON列取自错误的表格）：

SELECT stat.bug_num, stat.build_id
FROM bug_status stat
LEFT JOIN bug_status stat2
ON stat.closed_to = stat2.bug_num
WHERE stat.build_id = 3 OR stat2.build_id = 3

Answer 2

SELECT stat.bug_num,
       stat.build_id
  FROM bug_status  stat
 WHERE stat.build_id = 3
    OR stat.closed_to IN
        ( SELECT stat2.bug_num
            FROM bug_status  stat2
           WHERE stat2.build_id = 3
        )
;

（也可以通过JOIN或JOIN和UNION来实现，但我相信以上是最直观的方法。）

编辑添加：这是一个MySQL成绩单，展示了上述内容：

mysql> create table bug_status
    -> (bug_num numeric, build_id numeric, closed_to numeric);
Query OK, 0 rows affected (0.01 sec)

mysql> insert into bug_status values (1, 3, null);
Query OK, 1 row affected (0.00 sec)

mysql> insert into bug_status values (2, 4, null);
Query OK, 1 row affected (0.01 sec)

mysql> insert into bug_status values (3, null, 1);
Query OK, 1 row affected (0.00 sec)

mysql> insert into bug_status values (4, 3, null);
Query OK, 1 row affected (0.00 sec)

mysql> insert into bug_status values (5, null, 2);
Query OK, 1 row affected (0.00 sec)

mysql> SELECT stat.bug_num,
    ->        stat.build_id
    ->   FROM bug_status  stat
    ->  WHERE stat.build_id = 3
    ->     OR stat.closed_to IN
    ->         ( SELECT stat2.bug_num
    ->             FROM bug_status  stat2
    ->            WHERE stat2.build_id = 3
    ->         )
    -> ;
+---------+----------+
| bug_num | build_id |
+---------+----------+
|       1 |        3 |
|       3 |     NULL |
|       4 |        3 |
+---------+----------+
3 rows in set (0.00 sec)

已编辑添加，因为IN (...)方法在OP版本的MySQL中似乎不起作用：以下是另一种提供相同结果的查询：

SELECT stat.bug_num,
       stat.build_id
  FROM bug_status  stat
  LEFT
 OUTER
  JOIN bug_status  stat2
    ON stat.closed_to = stat2.bug_num
 WHERE stat.build_id = 3
    OR stat2.build_id = 3
;

Answer 3

为什么不：

DECLARE  @build_id int = <the build id>

SELECT   stat.bug_num, stat.build_id, stat.closed_to
FROM     bug_status stat
WHERE    stat.build_id = @build_id 
   OR    stat.closed_to = @build_id

？

Answer 4

此查询...

SELECT *
FROM bug_status t1
WHERE
    build_id = 3
    OR EXISTS (
        SELECT *
        FROM bug_status t2
        WHERE
            t2.build_id = 3
            AND t1.closed_to = t2.bug_num
    )

...产生以下结果：

bug_num   build_id    closed_to
1         3           NULL
3         NULL        1
4         3           NULL

用简单的英语：选择行：

build_id = 3
或与build_id = 3相关联的行。

sql自我加入多个属性

4 个答案: