UPDATE SQL柱選擇值

我有一個在列出對應於信貸員數字的表名prequals_created命名loanofficer列。我想更新此列有信貸員名稱，而不是數量，它位於一個名爲appusers表，其中fullname與loanofficer ID匹配。我試圖這樣做，首先使用SELECT通過使用名爲loan_pairing的「中間」表來生成信貸員名稱和id作爲一對錶。然而，我的語法有些問題。這裏是我的代碼：UPDATE SQL柱選擇值

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語句中的SELECT語句是正確的，並返回各自的ID成對信貸員的姓名。

這裏關於我的語法到底是什麼？

來源

2017-04-11 Jodo1992

是位於您的全名列在用戶？ –

是的，它是（額外字符） – Jodo1992

我覺得loan_pairing表亙古不抱什麼特別做這個工作，即APPUSER的id是prequals_created表作爲loanofficer，你想用全名是在appsuser表 –

在多表更新語句MySQL聯接的SET前走。我想，這應該爲你工作：

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;

來源

2017-04-11 16:23:17 GarethD

謝謝你做的，這個工作奇妙，我只需要在修改後還包括WHERE語句 – 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 
;

來源

2017-04-11 16:26:35 Teja

刪除不必要的連接表loan_pairing，並讓您的查詢有效和高效這是你想要的

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

來源

2017-04-11 16:27:48

UPDATE SQL柱選擇值

回答

相關問題