来自MySQL的错误消息

这是我的密码。我不确定我用它做错了什么。数据库名为inventory，表名为data。我试图在一个名为check_up date的字段中从最早的日期开始订购，并提取前两个最早的日期。我想排除名为“nobody”、“mainconf”和“testing”的用户：

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')

以下是错误：

ORDER BY

必须在

WHERE

子句之后和

限制之前

您不需要每次都重复

WHERE

。只要写下

和

，你就完成了

---

按排序的时间较晚（在之后，其中在限制之前）：

---

WHERE

子句需要位于

orderby

子句之前。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

---

正确的MySQL语法应该是：

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后面不需要“WHERE”，所有WHERE子句后面都需要ORDER BY

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

---

试试这个。

第一部分正确，但不能重复！（1064，'您的SQL语法有错误；请查看与您的MySQL服务器版本对应的手册，以了解在'WHERE data.name！=“mainconf”和'WHERE data.name！=“testing”order by checkup \'第1行附近使用的正确语法）@AndrewAlexander您真的读过我的答案了吗？在第二点中，我断言你必须有一个

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

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

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