How to get Html Source Code of A url in Php

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Hi , I need to display html source code of a url , how can i get it .

I need raw html code like this :

HTML

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">

<html xmlns="http://www.w3.org/1999/xhtml" >
<head id="Head1"><title>

</title></head>
<body>

I used it :

PHP

<?php  $data = file_get_contents("http://www.lorempixel.com/",0);
echo $data;
?>

But It was not getting Source code ,

Any help is appreciated !

Posted 28-Nov-14 23:07pm

Srinivas08

Updated 28-Nov-14 23:17pm

v3

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Comments

Zoltán Zörgő 29-Nov-14 7:16am

What have you got than?

Srinivas08 29-Nov-14 8:37am

simply all that html page itself was duplicated in my webpage

3 solutions

Solution 3

thanks , Now its working ... I am getting html source code as output !

Posted 29-Nov-14 2:35am

Srinivas08

Comments

Peter Leow 29-Nov-14 20:37pm

This is not a solution, should remove it. You can acknowledge the right solution by using the "Have a Question or Comment?" button. Do not forget to accept the right solutions.

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DamithSL · Accepted Answer · 2014-11-28T23:31:00

Solution 1

PHP

$data = file_get_contents('http://www.lorempixel.com/');
echo htmlspecialchars($data);

Posted 28-Nov-14 23:31pm

DamithSL

Peter Leow · Accepted Answer · 2014-11-28T23:44:00

Solution 2

Try this:

PHP

<?php  $data = file_get_contents("http://www.lorempixel.com/");
$html_encoded = htmlentities($data);
echo $html_encoded;
?>

Posted 28-Nov-14 23:44pm

Peter Leow

Updated 28-Nov-14 23:45pm

v2