sql问题选择查询 [英] Problem with sql Select Query

问题描述

我遇到此查询加入查询的问题



我收到错误，例如

您的SQL语法中有错误; 检查与 对应的手册您的MySQL服务器版本  正确语法  使用 near ' 其中missinginfo.missingid ='  31  ' '行 2  

我正在使用此查询

 $ result = mysql_query（  select * From missinginfo INNER JOIN adpersoncontact 
 ON missinginfo.missingid = adpersoncontact.missingid;其中missinginfo.missingid ='$ id'）或 die（mysql_error（））; 
 ？>  

解决方案

result = mysql_query（< span class =code-string> select * from missinginfo INNER JOIN adpersoncontact
ON missinginfo.missingid = adpersoncontact.missingid;其中missinginfo.missingid ='

id'）或 die（mysql_error（））;
？>

删除分号; 。如果 missingid 的类型是整数，则不需要引号。



另外，因为放了

I am getting problem with this query join query

I am getting an error like

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 missinginfo.missingid ='31'' at line 2

I am using this query

$result=mysql_query("select * From missinginfo INNER JOIN adpersoncontact
ON missinginfo.missingid=adpersoncontact.missingid; where missinginfo.missingid ='$id'  ")  or die (mysql_error());
?>

解决方案

result=mysql_query("select * From missinginfo INNER JOIN adpersoncontact ON missinginfo.missingid=adpersoncontact.missingid; where missinginfo.missingid ='

id' ") or die (mysql_error()); ?>

Remove the semicolon ;. And if the type of missingid is an integer, quotes are not necessary.

Also, because of putting the




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