Passing a variable from jquery to php

Question

I am trying to pass a variable from jquery to php

My jquery code:

<script>
$(window).load(function(){
$.get("as.php",url="http://www.wired.co.uk/news/archive/2013-01/10/chinese-desert-mystery", function(data,status){
$('#lax').html(data);
});
});
</script>
<div id="lax" >
</div>

And my "as.php" file is as follows:

<?php
include('phpQuery.php');
$file = mysqli_real_escape_string($dbc, strip_tags(trim($_GET['url'])));  
phpQuery::newDocumentFileHTML($file);
$titleElement = pq('title'); 
$title = $titleElement->html();
echo '<a href="" >' . htmlentities( $title) . '</a><br/>';

foreach(pq('meta') as $li)

  if ((pq($li)->attr('name')=='description')||(pq($li)->attr('name')=='Description')){
   echo '<p>'.pq($li)->attr('content').'</p>';
}
?>

I am tryin to pass 'url' variable from jquery code to my "as.php" file , but not able to do so. Where must be I going wrong?

btw you shouldn't wait for window load, use DOM ready function. $(document).ready(fn) — gdoron
– gdoron, Commented Jan 10, 2013 at 21:25
Do you have error handling in your database code? You should try to return just $_GET['url'] to see if the problem is in javascript or php. — jeroen
– jeroen, Commented Jan 10, 2013 at 21:28
My jquery is little bad.. please tell me how it is useful than window load function. — Robbie Dc
– Robbie Dc, Commented Jan 10, 2013 at 21:29

gdoron · Accepted Answer · 2013-01-10 21:20:55Z

4

You need to create an object

url="http://www.wired.co.uk/news/archive/2013-01/10/chinese-desert-mystery"

Should be:

{url :"http://www.wired.co.uk/news/archive/2013-01/10/chinese-desert-mystery"}

jQuery docs

answered Jan 10, 2013 at 21:20

gdoron

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2 Comments

gdoron Over a year ago

@RobbieDc, My PHP knowledge is about... zero and if you tried that already then your problem is in your PHP code.

Robbie Dc Over a year ago

Ok, Thanx for helping in the jquery part

jeroen · Accepted Answer · 2013-01-10 21:39:57Z

1

I don't see you opening a database connection, so with the code you posted $dbc will be NULL.

That causes mysqli_real_escape_string to return NULL as well.

As you are not doing any database operations, you should get rid of mysqli_real_escape_string completely.

answered Jan 10, 2013 at 21:39

jeroen

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5 Comments

Robbie Dc Over a year ago

Tried opening db connection, did not work. Any other alternative?

jeroen Over a year ago

@Robbie Dc Just remove it, you don't need it.

Robbie Dc Over a year ago

removed, now what to do next?

jeroen Over a year ago

@Robbie Dc That depends on the problem you are having :-) I would just open as.php in the browser with the required query string and debug line by line.

Robbie Dc Over a year ago

My problem is am not able to pass the 'url' variable to php code, do i use $.ajax instead of $.get ?

mamdouh alramadan · Accepted Answer · 2013-01-10 22:05:21Z

in you jquery:

<script>
$(window).load(function(){
$.get("as.php",
{url:"http://www.wired.co.uk/news/archive/2013-01/10/chinese-desert-mystery"}, function(data){
$('#lax').html(data);
});
});
</script>

in your php try using $_REQUEST

<?php
include('phpQuery.php');
$file = mysqli_real_escape_string($dbc, strip_tags(trim($_REQUEST['url'])));  
phpQuery::newDocumentFileHTML($file);
$titleElement = pq('title'); 
$title = $titleElement->html();
echo '<a href="" >' . htmlentities( $title) . '</a><br/>';

foreach(pq('meta') as $li)

  if ((pq($li)->attr('name')=='description')||(pq($li)->attr('name')=='Description')){
   echo '<p>'.pq($li)->attr('content').'</p>';
}
?>

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