Use PHP to include an HTML file as a jQuery object

Question

Let's say I have a "test.html" file with some example content:

<table>
 <tr>
  <td>'Test'</td>
  <td>Test "2"</td>
 </tr>
</table>

And I want to use that as a jQuery object. My first instinct would be to attempt:

var $testobject = $("<?php include('./test.html');?>");

But with the line breaks and quotes in test.html, this would fail. I specifically need the test.html file to be a jQuery object so it is hidden on load and is put in different places on the page with various triggered scripts, and I would prefer to use the more static PHP approach than a jQuery load. There must be a simple trick to manage this...

You could simply load the object into a variable using the jQuery ajax load function — random_user_name
– random_user_name, Commented Feb 25, 2012 at 14:51

Rob W · Accepted Answer · 2012-02-25 14:52:19Z

4

Read the static file using file_get_contents, and deal with special characters using json_encode:

var $testobject = $(<?php echo json_encode(file_get_contents('./test.html'));?>);

For example, if test.html contains:

<a href="/">
  Test
</a>

Then the output will be:

var $testobject = $("<a href=\"\/\">\n  Test\n<\/a>\n");

answered Feb 25, 2012 at 14:52

Rob W

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6 Comments

user173342 Over a year ago

Using my example test.html, this fails with this test.php file (nothing appears): <html><body> <script> var $testthing = <?php echo json_encode(file_get_contents('./test.html'));?>; $testthing.appendTo(document.body); </script> </body></html>

user173342 Over a year ago

I should add that the PHP echo succeeds, the HTML appears as you describe it. But it doesn't display from appending to the document body...

Rob W Over a year ago

@user You're missing the jQuery wrapper. $( <?php ... ?> );.

user173342 Over a year ago

Oops, added jQuery wrapper and still nothing: var $testthing = $(<?php echo json_encode(file_get_contents('./test.html'));?>);

Rob W Over a year ago

@user173342 You did never load the jQuery library.. See pastebin.com/jEpKRqq1 for the corrected version. (PHP: pastebin.com/RjN7H8sP).

|

Mouna Cheikhna · Accepted Answer · 2012-02-25 14:54:22Z

0

use load :

create a div

$('#test').load('test.html');

var testObject = $('#div').find('table');

answered Feb 25, 2012 at 14:54

Mouna Cheikhna

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Comments

Yaron U. · Accepted Answer · 2012-02-25 15:00:02Z

0

use output buffering:

<?php
ob_start();
include('./test.html');
$data = ob_get_contents();
ob_end_clean();
?>

var $testobject = $("<?=$data;?>");

answered Feb 25, 2012 at 15:00

Yaron U.

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