Question

我正在进行插入查询，如果已存在唯一键，则需要将大多数列更新为新值。它是这样的：

INSERT INTO lee(exp_id, created_by, 
                location, animal, 
                starttime, endtime, entct, 
                inact, inadur, inadist, 
                smlct, smldur, smldist, 
                larct, lardur, lardist, 
                emptyct, emptydur)
SELECT id, uid, t.location, t.animal, t.starttime, t.endtime, t.entct, 
       t.inact, t.inadur, t.inadist, 
       t.smlct, t.smldur, t.smldist, 
       t.larct, t.lardur, t.lardist, 
       t.emptyct, t.emptydur 
FROM tmp t WHERE uid=x
ON DUPLICATE KEY UPDATE ...; 
//update all fields to values from SELECT, 
//       except for exp_id, created_by, location, animal, 
//       starttime, endtime

我不确定UPDATE子句的语法是什么。如何引用SELECT子句中的当前行？

Answer 1

MySQL将假定equals之前的部分引用INSERT INTO子句中指定的列，第二部分引用SELECT列。

INSERT INTO lee(exp_id, created_by, location, animal, starttime, endtime, entct, 
                inact, inadur, inadist, 
                smlct, smldur, smldist, 
                larct, lardur, lardist, 
                emptyct, emptydur)
SELECT id, uid, t.location, t.animal, t.starttime, t.endtime, t.entct, 
       t.inact, t.inadur, t.inadist, 
       t.smlct, t.smldur, t.smldist, 
       t.larct, t.lardur, t.lardist, 
       t.emptyct, t.emptydur 
FROM tmp t WHERE uid=x
ON DUPLICATE KEY UPDATE entct=t.entct, inact=t.inact, ...

Answer 2

虽然我已经很晚了但是在看到那些想要INSERT-SELECT查询GROUP BY条款的人看到一些合理的问题之后，我想出了解决这个问题的方法。

进一步了解 Marcus Adams 和会计GROUP BY的答案，这就是我使用Subqueries in the FROM Clause

解决问题的方法

INSERT INTO lee(exp_id, created_by, location, animal, starttime, endtime, entct, 
                inact, inadur, inadist, 
                smlct, smldur, smldist, 
                larct, lardur, lardist, 
                emptyct, emptydur)
SELECT sb.id, uid, sb.location, sb.animal, sb.starttime, sb.endtime, sb.entct, 
       sb.inact, sb.inadur, sb.inadist, 
       sb.smlct, sb.smldur, sb.smldist, 
       sb.larct, sb.lardur, sb.lardist, 
       sb.emptyct, sb.emptydur
FROM
(SELECT id, uid, location, animal, starttime, endtime, entct, 
       inact, inadur, inadist, 
       smlct, smldur, smldist, 
       larct, lardur, lardist, 
       emptyct, emptydur 
FROM tmp WHERE uid=x
GROUP BY location) as sb
ON DUPLICATE KEY UPDATE entct=sb.entct, inact=sb.inact, ...

Answer 3

当 SELECT 语句有一个 GROUP BY 子句时。

    ....
    ON DUPLICATE KEY UPDATE    
    larct=VALUES(larct), lardur=VALUES(lardur),lardist= 
    VALUES(lardist)

INSERT INTO ... SELECT FROM ... on DUPLICATE KEY UPDATE

3 个答案: