Question

编辑 - 解决方案：

SELECT a.content as title, c.content as title_text, b.content as content FROM content_big AS b LEFT JOIN content_small AS a on a.id = 7 LEFT JOIN content_small AS c on c.id = 8 WHERE b.id = 1

我正在尝试做这样的选择：

从表content_big 获取列内容，其中ID = 1
从表content_small 获取列内容，其中ID = 7
从表content_small 获取列内容，其中ID = 8

可以使用1选择来获取这些数据吗？

我正在尝试此查询的一些变体：

 SELECT content_small.content as title, content_small.content as title, content_big.content as content
 FROM content_big LEFT JOIN
      content_small
      on content_small.id = 7 LEFT JOIN
      content_small
      on content_small.id = 8
 WHERE content_small.content = 1

但是我收到了这个错误

不唯一的表/别名：'content_small'

Answer 1

在联接上对表进行别名，如下所示：

SELECT cs1.content as title, cs1.content as title, content_big.content as content
 FROM content_big LEFT JOIN
      content_small cs1
      on content_small.id = 7 LEFT JOIN
      content_small cs2
      on content_small.id = 8
 WHERE cs2.content = 1

然后你就不会有命名冲突。

MYSQL选择左连接 - 如何使用2个差异来从同一个表中获取2行？

1 个答案: