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mysql_fetch_array(): supplied argument is not a valid MySQL result resource解决方法

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Release: 2016-06-13 13:39:27
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mysql_fetch_array(): supplied argument is not a valid MySQL result resource
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in D:\myapacheweb\test\self_info.php on line 7

PHP code
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mysql_connect("localhost","root","1"); 
    $query="select id from computer where user ='test2'";
    $result=mysql_query($query);
    
    echo "test2 使用了<br>";
    while($row = mysql_fetch_array($result))
    {
        echo $row['id']."开始时间".$row[starttime]."<br>";
    }

Copy after login

求助啊

------解决方案--------------------
$query="select id from computer where user ='test2'";

echo 一下$query 

把得到的SQL语句放在数据库里去执行一下,看能有没有数据。

这种问题一般都是 
SQL语句有错误,
或者数据库里没有这个数据造成的


------解决方案--------------------
SQL语句错了!
------解决方案--------------------
我一般是把
$result=mysql_query($query);
写成:
$result=mysql_query($query,$conn);


------解决方案--------------------
你没有指定数据库吧?- -
$result=mysql_query($query);
这个改成
$result=mysql_db_query('数据库名',$query);

------解决方案--------------------
探讨

我一般是把
$result=mysql_query($query);
写成:
$result=mysql_query($query,$conn);
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