Question

通过使用CASE语句更新我的SQL表，我遇到了问题。

SELECT number,amount,minimuminventory FROM artikel WHERE number=17;

+--------+--------+------------------+
| number | amount | minimuminventory |
+--------+--------+------------------+
|     17 |     10 |                0 |
+--------+--------+------------------+

我的金额为10但是当我更新我的桌子时：

UPDATE artikel 
SET amount = CASE WHEN amount - minimuminventory - 2 < 0 
                  THEN amount=amount-2 
                  ELSE amount=99 
             END 
WHERE artnr=17;

查询OK，1行受影响（0,01秒）
匹配的行数：1已更改：1警告：0

mysql将我的表更新为0

SELECT number,amount,minimuminventory FROM artikel WHERE number=17;

+--------+--------+------------------+
| number | amount | minimuminventory |
+--------+--------+------------------+
|     17 |      0 |                0 |
+--------+--------+------------------+

你看到了问题吗？

Answer 1

amount = CASE WHEN amount - minimuminventory - 2 < 0 THEN amount=amount-2 ELSE amount=99 END

CASE语句中的值是amount列的预期值，但您在其THEN和ELSE部分内再次执行amount=x。将其更改为：

amount = CASE WHEN amount - minimuminventory - 2 < 0 THEN amount-2 ELSE 99 END

Answer 2

你看到了问题吗？

在你的陈述中：

UPDATE artikel 
SET amount = CASE WHEN amount - minimuminventory - 2 < 0 
                  THEN amount=amount-2 
                  ELSE amount=99 
             END 
WHERE artnr=17;

比较被解释为布尔值而不是赋值。

SET amount = (amount=amount-2) -- comparison as if amount is equal to (17=17-2)
SET amount = (amount=99) -- comparison as if amount is equal to (17=99)
SET amount = false       -- i.e. 0, because 17 != 15 or 17 != 99

这导致0列中的值amount。

更改您的更新声明，如下所示：

UPDATE artikel 
SET amount = CASE WHEN amount - minimuminventory - 2 < 0 
                  THEN amount-2 
                  ELSE 99 
             END 
WHERE artnr=17;

MySQL Update CASE更新字段为＆＃34; 0＆＃34;

2 个答案: