Question

我想在查询中找到结果。问题是我使用IF语句根据该条目的另一列的值来对结果进行别名化。

这很好用

SELECT town
IF(is_found = 0, name, null) as not_found,
IF(is_found = 1, name, null) as found
FROM listing

但是如果我想按城镇分组我的结果，我需要连接名单列表的结果。

我试过了：

SELECT town
GROUP_CONCAT(IF(is_found = 0, name, null) as not_found) as missing_list,
GROUP_CONCAT(IF(is_found = 1, name, null) as found) as found_list
FROM listing

我收到了语法错误。

如何生成这样的连续列表，其中结果以IF语句为条件。

Answer 1

as不会进入if。所以你可以这样做：

SELECT town, 
       GROUP_CONCAT(IF(is_found = 0, name, null)) as missing_list,
       GROUP_CONCAT(IF(is_found = 1, name, null)) as found_list
FROM listing
GROUP BY town;

我更喜欢case而不是if（因为它几乎在所有数据库中都是标准的）：

SELECT town, 
       GROUP_CONCAT(CASE WHEN is_found = 0 THEN name END) as missing_list,
       GROUP_CONCAT(CASE WHEN is_found = 1 THEN name END) as found_list
FROM listing
GROUP BY town;

你可以GROUP_CONCAT IF结果为SELECT语句

1 个答案: