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Home/ Questions/Q 7606083
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Editorial Team
  • 0
Editorial Team
Asked: May 31, 20262026-05-31T00:19:21+00:00 2026-05-31T00:19:21+00:00

i am still having the same issue as before but this one is more

  • 0

i am still having the same issue as before but this one is more specified and more info here’s my form:

   <form method="post"action="">
   <table>
    <tr> <td>Login</td></tr>
    <tr><td> <input name="LEmail" id="LEmail" value="Email"/></td>
   <td> <label class="error" id="LEmail_error">*invalid</label></td>
   </tr>
   <tr>
   <td><input name="LPassword"  id="LPassword" value="Password"/></td>
   <td><label class="error" id="LPassword_error">*invalid</label>
  </td>
  </tr>
   </table>
  <input type="button" class="LButton" id="LButton" value="submit"/>
   </form>

and here is the jquery

$(document).ready(function(){
    $('.error').hide(); 
    $(".LButton").click(function(){
        //alert("yay i was clicked");
        var LEmail=$("input#LEmail").val()
        if(LEmail==""||LEmail=="Email"){
            $("label#LEmail_error").show(); 
            return false; 
        }
        var LPassword=$("input#LPassword").val(); 
        if(LPassword==""||LPassword=="Password"){
            $("label#LPassword_error").show();
            return false; 
        }
        var info='LEmail='+ LEmail + '&LPassword=' + LPassword;
        alert(info);
        $.ajax({
            type:"POST",
            url:"process.php",
            //dataType: string; 
            data: info,

            success:function(data){
                window.location.href="process.php";
                alert(data);
            },
            error: function(xhr, type, exception){
            alert("something went wront here"); 
            alert("Error: " + type);
            console.log('@Error: '+errorThrown);
            console.log('@Status: '+status);
            console.log('@Status Text: '+xhr.statusText);
              }
        })
    });
});

and the php

   $LEmail=$_POST['LEmail']; 
    $LPassword=$_POST['LPassword'];
    /*
    if(mysql_num_rows(mysql_query("SELECT EmailAddress AND Password FROM Users WHERE EmailAddress=  $LEmail AND Password=$LPassword)>0){
    }
    */

    var_dump($LEmail);
    var_dump($LPassword);

for some reason when the php page loads it claims it is an undefined index LEMail same for LPassword

all my variables are NUM but when i do the alerts the values are correct

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-31T00:19:22+00:00Added an answer on May 31, 2026 at 12:19 am

    Use a url encode around the POST data values:

    var info='LEmail='+ encodeURIComponent(LEmail) + '&LPassword=' + encodeURIComponent(LPassword);
    

    If your POST data is not properly encoded, the PHP script will not be able to parse it.

    Edit

    I think this is the problem

            success:function(data){
                window.location.href="process.php";
                alert(data);
            },
    

    When this happens your process.php has already processed the ajax call. You’re calling it for a second time here, and obviously with a redirect like that there is no POST data and that is why you’re seeing undefined index errors.

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