Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result reso,该怎么解决
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Release: 2016-06-13 11:57:39
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Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result reso 为什么查询的内容没有显示
if(isset($_POST[search_key])){ $tj=$_POST[sel_tj]; $key=$_POST[search_key]; $sql=mysql_query("select * from bg_user where $tj='$key'",$_link); $result=mysql_fetch_array($sql); if($result!=false){ echo "<script>alert('对不起,该用户不存在!');</script>"; }else{ do{ ?>
用户管理
姓名:
性别:
=$result['sex']?>
邮箱:
QQ:
=$result['QQ']?>
个人主页:
=$result['homepage']?>
IP地址:
生日:
=$result['birthday']?>
现住城市:
=$result['city']?>
家庭住址:
=$result['addr']?>
}while($result=mysql_fetch_array($sql)); } } ?>
------解决方案-------------------- 查询失败了! $sql=mysql_query("select * from bg_user where $tj='$key'",$_link) or die(mysql_error()); ------解决方案-------------------- 逻辑写反了,if($result!=false)这个该循环输出内容的,你alert了。 ------解决方案-------------------- 没有名为“用户名”的列
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