内部服务器错误php jquery ajax脚本GoDaddy

时间:2015-04-15 06:52:25

标签: php jquery

jQuery AJAX(index.php)

$.ajax({url:"xyz.php",type:"POST",
           data: {'login': email,'password': pass},
           dataType: "json",

            success:function(r){                
                if(r["status"]==='success')window.location.href="abc.php"
                else {$("#ErrorMsg").fadeIn();$("#Loader").hide();$("#SendButton").fadeIn();}
                },
                beforeSend:function(){$("#SendButton").hide();$("#Loader").fadeIn();},
                error:function(x,e){
                    if(x.status==0){
                        alert('You are offline!!\n Please Check Your Network.');
                    }else if(x.status==404){
                        alert('Requested URL not found.');
                    }else if(x.status==500){
                        alert('Internel Server Error.');
                    }else if(e=='parsererror'){
                        alert('Error.\nParsing JSON Request failed.');
                    }else if(e=='timeout'){
                        alert('Request Time out.');
                    }else {
                        alert('Unknow Error.\n'+x.responseText);
                    }
                    $("#Loader").hide();$("#SendButton").fadeIn();
                }
            });

PHP(xyz.php)

$DataSet = mysql_fetch_array(mysql_query("select * from administration where ADMIN_EMAIL_ID='$_POST[email]'"));
    if(strcmp($DataSet[2],$_POST['pass'])==0)
    {
        session_start();
        $_SESSION['Aunthentication'] = "Yes";
         echo json_encode(array('status' => "success"));
        //echo"approved";
    }
    else echo json_encode(array('status' => "failed"));    

0 个答案:

没有答案