如何在mysql中添加动态列 [英] How to add dynamic column in mysql

问题描述

SELECT *，CAST当likemsg.msgid =''THEN 1来自msg LEFT JOIN likemsg ON msg.id = likemsg.msgid



错误显示：



错误

SQL查询：文档



SELECT *，CAST
WHEN likemsg.msgid =''

THEN 1

FROM msg

LEFT JOIN likemsg ON msg.id = likemsg。 msgstr

LIMIT 0,30



MySQL说：文档



＃ 1064 - 您的SQL语法出错;检查与你的MySQL服务器版本相对应的手册，以便在'whenmsg.msgid ='附近使用正确的语法。来自msg LEFT JOIN的THEN 1喜欢msms.id = likemsg.msgi'在第1行



我尝试了什么：



SELECT *，CAST当likemsg.msgid =''那么1来自msg LEFT JOIN likemsg ON msg.id = likemsg.msgid

SELECT *,CAST when likemsg.msgid = '' THEN 1 from msg LEFT JOIN likemsg ON msg.id=likemsg.msgid

Error Show :

Error
SQL query: Documentation

SELECT * , CAST
WHEN likemsg.msgid = ''
THEN 1
FROM msg
LEFT JOIN likemsg ON msg.id = likemsg.msgid
LIMIT 0 , 30

MySQL said: Documentation

#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 'when likemsg.msgid = '' THEN 1 from msg LEFT JOIN likemsg ON msg.id=likemsg.msgi' at line 1

What I have tried:

SELECT *,CAST when likemsg.msgid = '' THEN 1 from msg LEFT JOIN likemsg ON msg.id=likemsg.msgid

推荐答案

小修正 -

small correction -

SELECT *,
IF(likemsg.msgid = '' , 1, likemsg.msgid ) as msgid
from msg LEFT JOIN likemsg ON msg.id=likemsg.msgid

阅读这篇文章 - 非常有用



mysql - &＃39; IF&＃39;在&＃39; SELECT&＃39; statement - 根据列值选择输出值 - Stack Overflow [ ^ ]

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