Question

我在表名loanofficer中有一个名为prequals_created的列，列出了与该贷款人员相对应的数字。我想更新此列以获得贷款人员姓名而不是号码，该号码位于名为appusers的表格中，其中fullname与loanofficer ID匹配。我试图通过首先使用SELECT生成一个贷款人员姓名表和id作为一对，使用名为loan_pairing的“中间”表来完成此操作。但是，关于我的语法有些不对劲。这是我的代码：

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，并返回与其各自ID配对的贷款人员的姓名。

我的语法到底是什么原因？

Answer 1

在MySQL中的多表更新语句中，联接在SET之前。我认为这对你有用：

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;

Answer 2

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
;

Answer 3

删除表 loan_pairing 的不必要的连接，并使您的查询有效且高效这就是你想要的

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使用SELECT值更新SQL列

3 个答案: