Question

这是我的代码。我不确定我做错了什么。该数据库名为inventory，该表名为data。我试图从名为check_up date的字段中的最早日期开始订购，并且最早拉出前两个。我想排除名称为“nobody”，“mainconf”和“testing”的用户：

SELECT user
FROM data
ORDER by checkup_date
WHERE data.name != "nobody"
   AND WHERE data.name != "mainconf"
   AND WHERE data.name != "testing" ASC
LIMIT 2

这是我的错误：

(1064, 'You have an error in your SQL syntax; check the manual that
corresponds to your MySQL server version for the right syntax to use near
\'WHERE data.name != "nobody" AND WHERE data.name != "mainconf" AND WHERE
data.nam\' at line 1')

Answer 1

这些是错误：

ORDER BY必须在WHERE clausole之后和LIMIT
每次WHERE时都不需要重复。只需写下AND即可。

Answer 2

SELECT user 
from data 
WHERE name != "nobody" AND name != "mainconf" AND name != "testing"
ORDER by checkup_date  ASC 
limit 2

试试这个。

Answer 3

排序稍后（ WHERE 之后 LIMIT 之前）：

SELECT user from data WHERE data.name != "nobody" AND data.name != "mainconf" AND data.name != "testing" ORDER by checkup_date ASC limit 2

Answer 4

WHERE子句需要在ORDER BY子句之前。而且您只需要指定一次WHERE子句。

SELECT user from data WHERE data.name != "nobody" AND data.name != "mainconf" AND data.name != "testing" ORDER by checkup_date ASC limit 2

Answer 5

正确的MySQL语法是：

SELECT user from data  WHERE data.name != "nobody" AND data.name != "mainconf" AND data.name != "testing" ORDER by checkup_date ASC LIMIT 2

每个AND之后不需要“WHERE”，ORDER BY需要在所有WHERE子句之后。

Answer 6

这应该这样做

SELECT 
    user 
FROM 
    data  
WHERE 
    data.name != "nobody" AND data.name != "mainconf" AND data.name != "testing" 
ORDER BY 
    checkup_date ASC 
LIMIT 2

来自MySQL的错误消息

6 个答案: