UPDATE SQL柱选择值

我有一个在列出对应于信贷员数字的表名prequals_created命名loanofficer列。我想更新此列有信贷员名称，而不是数量，它位于一个名为appusers表，其中fullname与loanofficer ID匹配。我试图这样做，首先使用SELECT通过使用名为loan_pairing的“中间”表来生成信贷员名称和id作为一对表。然而，我的语法有些问题。这里是我的代码：UPDATE SQL柱选择值

UPDATE 
    prequals_created AS pc 
SET 
    pc.loanofficer = lonames.fullname 
FROM 
    (SELECT DISTINCT a.fullname, lp.loanofficer 
    FROM appusers AS a 
    JOIN loan_pairing AS lp 
     ON a.id = lp.loanofficer 
    JOIN prequals_created AS pc 
     ON lp.loanofficer = pc.loanofficer) AS lonames 
WHERE 
    pc.loanofficer = lonames.loanofficer

嵌套在FROM语句中的SELECT语句是正确的，并返回各自的ID成对信贷员的姓名。

这里关于我的语法到底是什么？

来源

2017-04-11 Jodo1992

是位于您的全名列在用户？ –

是的，它是（额外字符） – Jodo1992

我觉得loan_pairing表亘古不抱什么特别做这个工作，即APPUSER的id是prequals_created表作为loanofficer，你想用全名是在appsuser表 –

在多表更新语句MySQL联接的SET前走。我想，这应该为你工作：

UPDATE prequals_created AS pc 
     JOIN loan_pairing AS lp 
      ON lp.loanofficer = pc.loanofficer 
     JOIN appusers AS a 
      ON a.id = lp.loanofficer 
SET  pc.loanofficer = a.fullname;

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2017-04-11 16:23:17 GarethD

谢谢你做的，这个工作奇妙，我只需要在修改后还包括WHERE语句 – Jodo1992

UPDATE 
    prequals_created AS pc 
SET 
    pc.loanofficer = lonames.fullname 
FROM 
    appusers AS a 
    JOIN 
    loan_pairing AS lp 
     ON a.id = lp.loanofficer 
    JOIN prequals_created AS pc 
     ON lp.loanofficer = pc.loanofficer 
;

来源

2017-04-11 16:26:35 Teja

删除不必要的连接表loan_pairing，并让您的查询有效和高效这是你想要的

UPDATE 
     pc 
    FROM appusers AS a 
     JOIN prequals_created AS pc 
      ON a.id = pc.loanofficer 
SET 
     pc.loanofficer = a.fullname

来源

2017-04-11 16:27:48

UPDATE SQL柱选择值

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