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Asked: 2026-06-14T19:05:02+00:00

Possible Duplicate: Jquery post, response in new window i have a table html in

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Possible Duplicate:
Jquery post, response in new window

i have a table html in a page called results.php that looks like
GenID
ENSMUSG00000098791
ENSMUSG00000023441
ENSMUSG00000047431

results.php have this function

<script>

function contenidoCelda() 
  {
  var table= $('#tabla_results');
  cells = $('td');
 for (var i=0,len=cells.length; i<len; i++)
   {
    cells[i].onclick = function()
      {
      var formData2 = new FormData(document.getElementById("formulario"));
      formData2.append("gen_id",(this.innerHTML));
      $.ajax({
          type: "POST",
      url: "test.php",
      data: formData2,
      cache: false,
      processData: false, 
      contentType: false, 
      success: function(data)
      {
        alert(data);       
            window.open('test.php', '_blank');

      }
    });

       }
     }
   }

</script>

i whant to use the data i send in the file test.php, not to return this to results.php, use in test.php, to generate the content dinamycally.

this is test.php

<?php
$data = $_POST['gen_id'];
system("mkdir $data");
echo "Hola";
echo $data;
echo '<xmp>';var_dump($data);echo '</xmp>';
?>

<html>
  <link href="css/ui-lightness/jquery-ui-1.9.1.custom.css" rel="stylesheet">
  <script src="js/jquery-1.8.2.js"></script>
  <script src="js/jquery-ui-1.9.1.custom.js"></script>
<body>
<form>

<p id = "testing"> Test page </p>

<?php if($data == "ENSMUSG00000047751") {echo "Good";} else {echo "Bad";}   ?>

</form>
</body>

so here in test.php, it must show Good, if i click that genid in the table
but it show Bad, and returns Good to results.php

the test where i create a dir, whit the genid i click works fine

what i must do?

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1 Answer

  1. Editorial Team

    In your $.ajax call, try setting:

    $.ajax({
  type: "POST",
  url: "test.php",
  data: "testing123",
  cache: false,
  processData: false, 
  contentType: false, 
});

    And see if dir test123 is created, and echoed back.

    If not, then the problem is in these two lines, specifically with your definition of var formData2:

    var formData2 = new FormData($('#formulario')[0]);
formData2.append("gen_id",(this.innerHTML));

    Check for typing errors, such as the comma in place of the semi-colon:

    var table = document.getElementById('tabla_results'),  <-- THIS IS A COMMA. TYPO?
cells = table.getElementsByTagName('td');

    Also, you have a mix of jQuery and javascript. Why not standardize on jQuery? For example, the two lines above would be written like this in jQuery:

    var table = $('#tabla_results');
cells = $('td');

    Much less typing, yes?

    Also, I’m sure you have, but are you sure you’ve included a link to the jQuery library? Such as:

    <script src="http://code.jquery.com/jquery-1.8.2.js"></script>
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