Undefined index in php when using ajax

Question

My Form page :

<body>  
<head>
        <script src="http://code.jquery.com/jquery-1.9.1.js"></script>
        <script>
        function val()
        {
            var name=document.getElementById("name").value;
            if(name == '')
            {
                alert(name+" is empty");
            }
            else
            {
            var dataString = "name = "+name;
            $.ajax({
                type:"POST",
                url:"hi.php",
                data:dataString,
                cache:false,
                success:function(html){
                    $('#msg').html(html);
                }
            });
            }
        return false;
        }
        </script>

    </head>
<body>
    <form>
        <input type="text" id="name" >
        <br/><br/>
        <input type="submit" value ="submit" onclick="return val()">
    </form>
<p id="msg"></p>
</body>

Here is my hi.php file

<?php
$name = $_POST["name"];
echo "Response : ".$name;
?>

When clicking on submit button it show an error Notice: Undefined index: name in C:\wamp\www\SendEmailAjaxJquery\hi.php on line 2

I don't know where is the error plz help me to find out the error...

Thanks in advance

always try printing the received data from ajax before using it, so that you can be sure about what is being received on what indexes. — Pardeep Poria
– Pardeep Poria, Commented May 5, 2016 at 6:02
Possible duplicate of Undefined index: Error in ajax POST and/or php script? — Ani Menon
– Ani Menon, Commented May 5, 2016 at 6:03

Waqas_aamer · Accepted Answer · 2016-05-05 06:09:45Z

1

Use this in place of this

var dataString = "name = "+name;

change into this

var dataString = 'name='+ name;

i have the same case here you will be guided from it.

 <html>
 <head>
  <script type="text/javascript" src="jquery-1.12.3.min.js"></script>
 </head>
 <body>
 <div class="content">
    <input type="text" class="search" id="searchid" placeholder="Search for    people" />
    <div id="result"></div>
 </div>  
 </body> 
 </html>
 <script type="text/javascript">
 $(function(){
  $(".search").keyup(function() 
  { 
    var searchid = $(this).val();
    var dataString = 'search='+ searchid;
    if(searchid!='')
    {
      $.ajax({
      type: "POST",
      url: "result.php",
      data: dataString,
      cache: false,
      success: function(html)
      {
         $("#result").html(html).show();
      }
     });
   }return false;    
   });

edited May 5, 2016 at 6:09

answered May 5, 2016 at 6:04

Waqas_aamer

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3 Comments

Ani Menon Over a year ago

And how would that help?

Waqas_aamer Over a year ago

You can see it now @Ani Menon

Devraj Over a year ago

Thanks its working now... just silly mistake it was... really its helpful thanks again...

J.C. Fong · Accepted Answer · 2016-05-05 06:07:18Z

1

Change to:

$.ajax({
  type:"POST",
  url:"hi.php",
  data: {name: name},
  cache:false,
  success:function(html){
    $('#msg').html(html);
  }
});

jQuery ajax data setting accept Object to customize the key. For example:

data: {anything: "123"}

In PHP:

echo $_POST["anything"]; //123

answered May 5, 2016 at 6:07

J.C. Fong

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Comments

Maninderpreet Singh · Accepted Answer · 2016-05-06 02:44:03Z

0

in your ajax data change to

 data:{ 'name' : name }

edited May 6, 2016 at 2:44

answered May 5, 2016 at 6:00

Maninderpreet Singh

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