2017-04-17 141 views
0

嗨,当我把这个查询选择这个工作好,使输出在mysql中where命令不起作用

select SUBSTR(img_address1,LOCATE('/',img_address2)+1,36)fROM content 

但是当我做这个查询此做出错误

SELECT id from content WHERE ((SUBSTR(img_address2,LOCATE('/',img_address2)+1,5) LIKE'salam') 

和本作错误太

SELECT id from content WHERE ((SUBSTR(img_address2,LOCATE('/',img_address2)+1,5) = 'salam') 

和我的错误是:

SELECT id from content WHERE ((SUBSTR(img_address2,LOCATE('/',img_address2)+1,5) = 'salam') LIMIT 0, 25 
MySQL said: Documentation 

#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'LIMIT 0, 25' at line 1 

回答

0

SUBSTR之前有一个额外的开放支架(。删除它将修复错误。

SELECT id from content WHERE (SUBSTR(img_address2,LOCATE('/',img_address2)+1,5) = 'salam') LIMIT 0, 25 
+0

哦,谢谢我能找到它你的权利 – nazanin