请教一个 Mysql 条件更新语句,一句完成 - V2EX
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uti6770werty
V2EX    MySQL

请教一个 Mysql 条件更新语句,一句完成

  •  
  •   uti6770werty 2021-04-18 22:43:59 +08:00 2452 次点击
    这是一个创建于 1710 天前的主题,其中的信息可能已经有所发展或是发生改变。

    在同一个表中,对值的进行判断,根据判断结果[是否]再进行更新

    1 、有一个名叫mtt的表
    2 、mtt 表内有三个涉事字段,AID,mttID,UpdateTime
    3 、mttID='3A2NN179'的条目,如果AID不等于'2719',把这些条目的AID更新为'2719',同时更新 UpdateTime 时间为 now(),
    4 、如果AID已经是 2719,则不用理会,也不用更新 UpdateTime

    SELECT * FROM mtt WHERE mttID='3A2NN179';

    程序上来说,
    用 SELECT 到内容,程序本地判断,再去做两次 UPDATE,两三个交互的是能做到,握手延时等,费效比不好,看看有无一句完成的可能?

    4 条回复    2021-04-19 14:26:49 +08:00
    CEBBCAT
        1
    CEBBCAT  
       2021-04-18 22:53:27 +08:00   1
    UPDATE mtt SET AID='2719',UpdateTime=UNIX_TIMESTAMP() WHERE mttID='3A2NN179' AND AID!='2719'


    这样?
    uti6770werty
        2
    uti6770werty  
    OP
       2021-04-18 23:01:44 +08:00
    @CEBBCAT 赞一个,思路拐了一个弯就没想到,感谢!
    jorneyr
        3
    jorneyr  
       2021-04-19 09:45:51 +08:00
    UPDATE question_knowledge_point qkp
    JOIN (SELECT knowledge_point_id AS id, COUNT(id) AS count FROM question GROUP BY knowledge_point_id) AS t ON qkp.id=t.id
    SET qkp.count=t.count
    devswork
        4
    devswork  
       2021-04-19 14:26:49 +08:00
    update xxx where mttID='3A2NN179' and AID != '2719'
    不就可以啦?
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