查询作为SELECT而不是更新？

Question

我试图在MySQL中使用联系人表和accounts_contacts连接表执行重复的清理查询。我有查询工作作为一个SELECT查询，但当我尝试使其更新，我得到一个非特定的错误：

＃1064 - 您的SQL语法中有错误;检查对应于你的MySQL服务器版本使用附近的正确语法手册“FROM sugarDB.contacts INNER JOIN（SELECT dupIDs.id FROM（SELECT克拉”在行3

下面是该查询：

UPDATE ctUpdate 
    SET ctUpdate.deleted = 1 
    FROM sugarDB.contacts AS ctUpdate 
    INNER JOIN (
     SELECT dupIDs.id 
     FROM (
      SELECT ctIDs.id 
      FROM sugarDB.contacts AS ctIDs 
      INNER JOIN (
       SELECT ctSource.first_name, 
        ctSource.last_name 
       FROM sugarDB.contacts AS ctSource 
       GROUP BY ctSource.first_name, 
        ctSource.last_name 
       HAVING COUNT(*) > 1 
       ORDER BY COUNT(*) DESC 
      ) 
      AS ctSource 
      ON ctIDs.first_name = ctSource.first_name 
       AND ctIDs.last_name = ctSource.last_name 
     ) 
     AS dupIDs 
     LEFT JOIN sugarDB.accounts_contacts AS a2cIDs 
     ON dupIDs.id = a2cIDs.contact_id 
     WHERE a2cIDs.id IS NULL 
    ) 
    AS dupIDs 
    ON ctUpdate .id = dupIDs.id 
; 

我已经倒在这几天了，我找不到任何错误的帮助是极大的赞赏

Answer 1

There is no FROM clause in UPDATE statements in MySql.

相反，你的加入应该是你UPDATE条款的一部分。！

UPDATE sugarDB.contacts AS ctUpdate 
    INNER JOIN (
     SELECT dupIDs.id 
     FROM (
      SELECT ctIDs.id 
      FROM sugarDB.contacts AS ctIDs 
      INNER JOIN (
       SELECT ctSource.first_name, 
        ctSource.last_name 
       FROM sugarDB.contacts AS ctSource 
       GROUP BY ctSource.first_name, 
        ctSource.last_name 
       HAVING COUNT(*) > 1 
       ORDER BY COUNT(*) DESC 
      ) 
      AS ctSource 
      ON ctIDs.first_name = ctSource.first_name 
       AND ctIDs.last_name = ctSource.last_name 
     ) 
     AS dupIDs 
     LEFT JOIN sugarDB.accounts_contacts AS a2cIDs 
     ON dupIDs.id = a2cIDs.contact_id 
     WHERE a2cIDs.id IS NULL 
    ) 
    AS dupIDs 
    ON ctUpdate .id = dupIDs.id 
    SET ctUpdate.deleted = 1 
;