Question

我正在使用JQuery Validation Plugin。我看上去处理违约的营地档案的边远功能。

I modified the php file to use my own version but mysql is returning

Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in    /home/fastbluf/syatch/module/1.func.php on line 15

My PHP 《刑法》如下。我的所有辛子都看得正确。

<?php
// Last Edited: 4/23/12
$conn = mysql_connect( localhost , hidden , hidden ) or die( Iam dying );
$rs = @mysql_select_db( "hidden", $conn) or die( "Err:Db" );

$do = $_REQUEST[ do ];
$email= $_REQUEST[ email ];
$user= $_REQUEST[ user ];

function checkInfo($do,$email,$user){
    switch ($do) {
        case 1:
            $sql = "select * from User_Base where Email_Address = $email";
            $results = mysql_query($sql). mysql_error();
            $nResults = mysql_num_rows($results);
            if ($nResults > 0) {
                $valid="false";
            } else {
                $valid="true";
            }
        break;

        case 2:
        //not yet 
        break;      
    }
    return $valid;
}
echo checkInfo($do,$email,$user);
?>

Answer 1

问题在于,你对你的结果重新思考,使结果不再有效。

$results = mysql_query($sql). mysql_error();

改变这种状况就是这样:

$results = mysql_query($sql) or die(mysql_error());

你的问询也应当改变,以引用电子邮件地址,为了防止攻击,应躲避地址(SQL注射):

$email = mysql_real_escape_string($_REQUEST[ email ]);
$sql = "select * from User_Base where Email_Address =  $email ";

Answer 2

女士们

$sql = "select * from User_Base where Email_Address =  ".$email." ";

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