在MySQL的COALESCE中选择

Question

我正在尝试在MySQL中执行以下操作:

UPDATE
    x
SET
    y = COALESCE(SELECT z FROM table WHERE a = b AND c = d AND e = f LIMIT 1,
                 SELECT z FROM table WHERE a = b AND c = d LIMIT 1,
                 SELECT z FROM table WHERE a = b LIMIT 1);

这对我来说听起来很有效.我正在尝试使用最合适的值更新列.如果我能找到符合3个标准的记录 - >那就是我需要的记录.另外一个适合2个标准,否则记录只匹配一个标准.

我可以在3个更新查询中执行此操作,但我不明白为什么这不起作用.根据手册:

COALESCE returns the first non-NULL value in the list, or NULL if there are no non-NULL values.

这正是我需要的.但它给出以下错误:

Error Code: 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 'SELECT'

我在这里错过了什么吗？

Answer 1

您的问题可能是您获得的子选择返回的z值超过1 z.你只能得到一个.

你可以尝试:

update x
set y = (select z from t where a = b
         order by (case when a = b and c = d and e = f then 1
                        when a = b and c = d then 2
                        else 3
                   end)
         limit 1
        );

但是,您的特定语法错误是您需要围绕选择括号.所以这是语法上正确的版本:

UPDATE
    x
SET
    y = COALESCE((SELECT z FROM table WHERE a = b AND c = d AND e = f LIMIT 1),
                 (SELECT z FROM table WHERE a = b AND c = d LIMIT 1),
                 (SELECT z FROM table WHERE a = b LIMIT 1));