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Home/ Questions/Q 6762015
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Editorial Team
  • 0
Editorial Team
Asked: May 26, 20262026-05-26T14:15:46+00:00 2026-05-26T14:15:46+00:00

I am sending this variable to other page, to make a validation. However print_r($_POST);

  • 0

I am sending this variable to other page, to make a validation.

However print_r($_POST); only show Array ().

Normally we use serialize, is sent the name attribute, but in this case only is sent the variable with the text. So something like $form = $_POST['data']; only works to

in firebug:

data="something"

but in my case i just send

something

code

 <?php $page = "someText"; ?>

JS

 default:
            $("#msg").fadeTo(200, 0.1, function() {
                $(this).html('success').fadeTo(900, 1);

                $.ajax({
                    url: "page_validation.php",
                    type: "post",
                    dataType: "json",
                    data:'<?php echo $page; ?>',          
                    success: function(data) {
                        $('#one').load('page.php');
                        }
                });
            });
            break;
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-26T14:15:46+00:00Added an answer on May 26, 2026 at 2:15 pm

    try it like

    $page = "someText";
    
     $.ajax({
       url: "page_validation.php",
       type: "post",
       dataType: "json",
       data:"postedvariable=<?php echo $page; ?>",       
       success: function(data) {
                $('#one').load('page.php'); 
       }
     });
    

    and acces the varible in page_validation.php as $_POST[‘postedvariable’]

    also another option will be to use

    $.post("page_validation.php",
    {postedvariable:'<?php echo $page; ?>'},
    function(data) {
      $('#one').load('page.php');
     });
    

    whic looks even simpler than the $.ajax

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