Question

我正在尝试构建一个查询，该查询将采用一些默认值并将保持与表的连接。这只是为了在主查询（第二个）与任何行不匹配的情况下允许至少一个结果，因此可能返回空结果集as mentioned on my another question。

我的查询大致如下：

SELECT * FROM(
    (SELECT 'test' as `column1`, 'test' as `column2`, 0 as c) t1
    LEFT JOIN
        (SELECT `column1`, `column2`, COUNT(*) as c
        FROM `my_table`
        WHERE `status` = 1
        GROUP BY column1, column2) t2
)

它在phpmyadmin上显示一个mysql错误，如下所示：

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 't2 )' at line 7

请问查询有什么问题？

Answer 1

我认为这是您想要的查询：

SELECT t1.column1, t2.column2, coalesce(t2.c, 0)
FROM (SELECT 'test' as `column1`, 'test' as `column2`, 0 as c
     ) t1 LEFT JOIN
     (SELECT `column1`, `column2`, COUNT(*) as c
      FROM `my_table`
      WHERE `status` = 1
      GROUP BY column1, column2
     ) t2
     on t1.column1 = t2.column1 and t1.column2 = t2.column2;

您的查询不完整。例如，它没有on子句。

左连接有一些默认值

1 个答案: