CODEWITHSUNDEEP
phpjavascriptajaxformajax-forms
user1583647
user1583647

Reputation: 1245

Return value from ajax form

I have a login form in php. I have used ajaxform so that the page does not get reloaded when the username or password is wrong. I have another page checklogin.php which checks whether the username and pw is there in the database and it returns the count of the number of rows. I want to compare the count in the login.php and display the error message if the count=1 and redirect to another page if count=2. I tried to display the count in an errordiv using target: 'errordiv' and checking its innerHTML but it failed to do so.

My login.php 

<script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.8/jquery.min.js"></script>
  <script type="text/javascript" src="scripts/jquery.form.js"></script>
<!--Slider-in icons-->
<script type="text/javascript">


    function checklogin()
    {
     var request=$("#login-form").ajaxForm(
                    {

                  target:'#errordiv'

                    }).abort();
                request.submit();
        var divmsg=document.getElementById("errordiv");
        if (divmsg.childNodes[0].nodeValue == "1")
            {
        divmsg.childNodesp[0].nodeValue="Sorry";
        alert ("Invalid username or password");} //I didn't get the alert as well a the divmsg content didn't change


                    };
</script>

</head>
<body>

<!--WRAPPER-->
<div id="wrapper">

<!--LOGIN FORM-->
<form name="login-form" id="login-form" class="login-form" action="checklogin.php" method="post">


    <!--CONTENT-->
    <div class="content">
    <!--USERNAME--><input name="un" type="text" class="input username" placeholder="Username" onfocus="this.value=''" required="required" /><!--END USERNAME-->
    <!--PASSWORD--><input name="pw" type="password" class="input password" placeholder="Password" onfocus="this.value=''" required="required" /><!--END PASSWORD-->
    </div>
    <!--END CONTENT-->

    <!--FOOTER-->
    <div class="footer">
    <!--LOGIN BUTTON--><input type="submit" name="submit" value="Login" class="button" onclick="checklogin()" /><!--END LOGIN BUTTON-->

    </div>
    <!--END FOOTER-->

</form>
<div id="errordiv" class="errordiv"></div>
<!--END LOGIN FORM-->

checklogin.php

<?php

 include ('dbconn.php');
       $user = trim($_POST['un']);
        $pass = trim($_POST['pw']);
        $user = mysql_real_escape_string($user);
        $pass = mysql_real_escape_string($pass);
       $result=mysql_query("select Id from login where Username='$user' and Password='$pass'") or die (mysql_error());
      $result= mysql_fetch_assoc($result);
     $num= count ($result);
    echo $num;
?>

How can I get the value of $num in login.php without displaying it to a div and compare the value of $num and accordingly display the errormsg in errordiv or if $num==2 redirect to another page.

Upvotes: 0

Views: 1648

Answers (2)

Chickenrice
Chickenrice

Reputation: 5727

Maybe try to make the checking request asynchronously and process user feedback in AJAX callback:

JS:

    $(function(){
        $("#login-form").ajaxForm(function(response){
            if(response.count===1){
                $("#errordiv").html("msg you want to show");
            }else if(response.count===2){
                //redirect
            }
        });
    });

php:

<?php
    header("Content-Type: application/json");

    //your database check logic
    $user = trim($_POST['un']);
    $pass = trim($_POST['pw']);
    $user = mysql_real_escape_string($user);
    $pass = mysql_real_escape_string($pass);
    $result=mysql_query("select Id from login where Username='$user' and Password='$pass'") or die (mysql_error());
    $result= mysql_fetch_assoc($result);
    $num= count ($result);

    //and return response in JSON
    echo json_encode(array("count"=>$num));

?>

Hope this is helpful for you.

[EDIT]

remove the form submit button inline JS function invoking:

onclick="checklogin()"

and put checklogin function logic to document ready callback, initialize the form when the DOM is ready

Upvotes: 0

DevZer0
DevZer0

Reputation: 13535

Change your code to include the code which change the div and provide alert to be included in the callback.

     var request=$("#login-form").ajaxForm(
                 {
                  target:'#errordiv', 
                  success: function (html) {
                      if ($("#errordiv).html() == "1") {
                          $("#errordiv).html("Sorry");
                          alert ("Invalid username or password");  
                      }
                  }
                    }).abort();
                request.submit();

Upvotes: 1

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