Question

这是一个挑战：我有三张桌子。我需要使用查询结果填充table_1。此查询的输入是table_1的id。我怎么能这样做？

谢谢。

我已经尝试了

UPDATE  table_1 AS a 
           INNER JOIN
           (
               SELECT  s.some_name 
               FROM    table_2 as st
               JOIN table_3 AS s ON s.a_id = st.a_id 
               ORDER BY st.somefiled DESC
               LIMIT 1 
           ) AS b ON  a.a_id = b.a_id AND a.t_id = b.t_id 
           SET     a.some_name = b.some_name

但是这给了我错误

1054 - 'on子句'中的未知列'b.a_id' 虽然，a_id存在于table_2和table_3
中

提前谢谢

Answer 1

您还必须在选择列表中显示Id列。像这样的事情： -

UPDATE  table_1 AS a 
       INNER JOIN
       (
           SELECT  s.some_name, st.a_id, st.t_id
           FROM    table_2 as st
           JOIN table_3 AS s ON s.a_id = st.a_id 
           ORDER BY st.somefiled DESC
           LIMIT 1 
       ) AS b ON  a.a_id = b.a_id AND a.t_id = b.t_id 
       SET     a.some_name = b.some_name

我不知道这是否能解决你的目的。但这将删除错误

MySQL的。通过子查询更新表中特定列的所有行，该子查询从每行获取输入值

1 个答案: