Undefined index error in php using ajax

Question

<form role="form" method="post" action="test.php">
<label for="contact">Mobile No:</label><br>
      <input type="tel" class="form-control" name="contact" title="Mobile number should not contain alphabets. Maxlength 10" placeholder="Enter your phone no" maxlength="15" required  id='contact_no'>
      <br><br>
<button type="submit" class="btn btn-success" name="submit" id="submit">Submit</button>
    <button type="reset" class="btn btn-default" id='reset'>Reset</button>
  </form>

Ajax and Javascript Code
script type="text/javascript">
    $(document).ready(function(){
        $("#submit").click(function(){
        var dialcode = $(".country-list .active").data().dialCode;
        var contact = $("#contact_no").val().replace(" ","");
        var countrycode = $('.country-list .active').data().countryCode;
            var cn;
            var cc;
            var dc;
            $.ajax({
            url: "test.php",
            type: "POST",
            data: {'cc' : contact},
            success: function(data) 
            {
                alert("success"); 
              }  
            });
        });
    });
    </script>

The variables show the values if displayed by alert message but are not passed on to the test.php page. It shows undefined index error at the following statement

test.php is as follows

<?php

    if(isset($_POST['submit'])){

         $contact = $_POST['cc'];       //it shows the error here

    }
    echo  $contact;

I had referred to many websites which show the same thing. It dosent work for me. I think the syntz of ajax is correct and have tried all possibilities but still dosent work. Please help

^this, you don't need quotes around your keys in a javascript array — Jeremy C.
– Jeremy C., Commented Jun 4, 2015 at 11:18
You're posting {cc: contact}, but you're checking for $_POST['submit'] which isn't being sent. The callback also doesn't stop the event, so you might want to return false (stops default and propagation) — Elias Van Ootegem
– Elias Van Ootegem, Commented Jun 4, 2015 at 11:23
@Naitik can you please try the codes, and mark anyone as answer which gives you correct result that you want and that will be descriptive also. you can vote-up others if they are helpful too. thanks. we diid code for you. at-least check them. — Death-is-the-real-truth
– Death-is-the-real-truth, Commented Jun 4, 2015 at 13:56

Elias Van Ootegem · Accepted Answer · 2015-06-04 11:29:04Z

2

You're posting {cc: contact}, but you're checking for $_POST['submit'] which isn't being sent. The callback also doesn't stop the event, so you might want to return false (stops default and propagation). Something like this should do the trick:

$('#submit').on('click', function()
{
    //do stuff
    $.ajax({
        data: {cc: contact},
        method: 'post',
        success: function()
        {
            //handle response here
        }
    });
    return false;
});

Then, in PHP:

if (isset($_POST['cc']))
{
    //ajax request with cc data
}

Also not that this:

$("#contact_no").val().replace(" ","");

Will only replace 1 space, not all of them, for that you'll need to use a regex with a g (for global) flag:

$("#contact_no").val().replace(/\s+/g,"");

answered Jun 4, 2015 at 11:29

Elias Van Ootegem

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Comments

Saty · Accepted Answer · 2015-06-04 11:28:41Z

0

You are using ajax to form submit

and you use $_POST['submit'] to check it would be $_POST['cc']

test.php

<?php
    if(isset($_POST['cc'])){// change submit to cc
      $contact = $_POST['cc'];//it shows the error here
    }
echo  $contact;

answered Jun 4, 2015 at 11:28

Saty

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Comments

Spinstaz · Accepted Answer · 2021-11-06 21:37:39Z

0

@Saty answer worked for me, but my code on ajax was a bit different. I had multiple form data wrapped up into a form variable, that was passed to the php page.

const form = new FormData();

form.append('keywords', keywords);
form.append('timescale', timescale);
form.append('pricing_entered', pricing_entered);

 $.ajax({
   url: "../config/save_status.php",
   data: form,
   method: "POST",
   datatype: "text",

success: function (response, data) {

                   }

Then my php was:

if (isset($_POST['data'])) {
 // all code about database uploading
}

answered Nov 6, 2021 at 21:37

Spinstaz

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