Issues in posting JS variable to PHP script

Question

I'm trying to post a JS variable to a PHP script but I'm not able to do it. Here are the scripts,

The following JS is embedded in a script named "check.php",

<script type="text/javascript">
$(document).ready(function() {
$('#content-area').mouseup(function() {
 var selection = getSelected();
  if(selection && (selection = new String(selection).replace(/^\s+|\s+$/g,''))) {
     alert('Sending the following text to the server via AJAX: ' + selection);
            $.ajax({
             type: 'post',
             url : 'calculate.php',
             data: 'selection=' + encodeURI(selection)  
            });
     }
   });
   });  
</script>

I'm posting the "selection" variable using ajax to The PHP script(calculate.php) where I'm receiving the value as follows.

<?php
    if(!empty($_SERVER['HTTP_X_REQUESTED_WITH']) &&
      strtolower($_SERVER['HTTP_X_REQUESTED_WITH']) == 'xmlhttprequest' &&  
      $selection = trim($_POST['selection'])) {
        $selText = $selection;
       }
?>

I'm getting a null value at the server-side i.e "calculate.php". Please tell me what is going wrong here?

Gowri · Accepted Answer · 2014-01-24 12:02:15Z

1

Try this once.

in ajax call

$.ajax({
            type: 'post',
            url : 'calculate.php',
            data: { selection : selection},
          });

In calculate.php get your data using $_POST['selection'];

edited Jan 24, 2014 at 12:02

answered Jan 24, 2014 at 11:54

Gowri

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2 Comments

kmario23 Over a year ago

I checked it using if(isset($_POST['selection']) in calculate.php, and the data is not received at the server side. But it shows the alert msg on selecting the text. Pl help

DontVoteMeDown Over a year ago

Why post an answer with same content/proposal as other already posted answer? If it didn't worked for that answer, why it should works for yours?

DontVoteMeDown · Accepted Answer · 2014-01-24 10:58:15Z

1

Change you data attribute to:

data: { selection: encodeURI(selection) }

Then it will be available on your PHP script as:

$_POST['selection']

Further, look at your if condition in your PHP script:

$selection = trim($_POST['selection'])

If you're want to compare those values, I think you should use ==. If not, just move it outside the condition.

edited Jan 24, 2014 at 10:58

answered Jan 24, 2014 at 10:44

DontVoteMeDown

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3 Comments

kmario23 Over a year ago

Still no success. Still getting null value. Do you have any other fix?

kmario23 Over a year ago

Even the updated thing is not working. Also I checked it using if(isset($_POST['selection']) in calculate.php, and the data is not received at the server side. But it shows the alert msg on selecting the text. I'm disheartened, Pl help

DontVoteMeDown Over a year ago

@mario23 is hard to help you with that lack of information. Post the result of print_r($_POST); at the beggining of your PHP script and your request from developers tools(Chrome) - it can be a screenshot or even request header text...

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