How to set jquery variable in PHP CURL

Question

I have a screen scrape code in PHP like this:

<?
$url = 'https://www.google.nl/search?q=cars';
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$curl_scraped_page = curl_exec($ch);
curl_close($ch);
echo $curl_scraped_page;
?>

I also have a Jquery(ajax) script which can fetch the proxy like this:

$.get('ThePhpFile.php', function(data){

     $(data).appendTo('div')

}

Everything is working fine, but now I would like to set the URL to be scraped as a variable which is in fact the value of some input in parent document (I'm working with iframes). I know how to do this in Jquery:

var TheUrl = $("input", parent.document.body).val();

So my question is how do I set the variable to work with the PHP code. Is it necessary to put it in the PHP code? How do I do that?

@mplungjan I already tried several things but with no luck(not good at PHP) If I Google I only encounter codes which go in the opposite direction PHP variable ----> Jquery — Youss
– Youss, Commented Mar 9, 2013 at 9:00

mplungjan · Accepted Answer · 2013-03-09 14:12:32Z

1

You need to ajax the variable to the php:

This will send ?url=..... to the php

Full example DEMO

<html>
<head>
<title>Search Proxy</title>
<script type='text/javascript' src='//code.jquery.com/jquery-1.9.1.js'></script>
<script>
$(function() {
  $("#search").on("submit",function(e){
    e.preventDefault();
    $.get('searchproxy.php', 
     {url:$("#url").val()}, 
       function(data){
         $(data).html('#result');
     });
  });
});
</script>
</head>
<body>
<div>
<form id="search">
<input id="url" type="text" value="" />
</form>
<div id="result"></div>
</body>
</html>

<?php
$url = $url = $_GET["url"]; // you need to add input cleaning
$ch = curl_init($url);
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
$curl_scraped_page = curl_exec($ch);
curl_close($ch);
echo $curl_scraped_page;
?>

edited Mar 9, 2013 at 14:12

answered Mar 9, 2013 at 8:58

mplungjan

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8 Comments

Youss Over a year ago

Im getting 'missing name in function' error when I put it in JsFiddle jsfiddle.net/4ZBEf

mplungjan Over a year ago

Also your fiddle has no html and is not running jQuery

Youss Over a year ago

But how do I relate this to the PHP code.. If I remove the URL part($url = 'google.nl/search?q=cars';) it doesn't work..

Youss Over a year ago

I mean something like this $url = TheUrl();

Youss Over a year ago

It doesn't work for me..Could the problem be parent.document.body).val() I would really appreciate help with this lat bit..

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