Get html source of the referring url

I have a php script that websites on the internet will use. I would like it so that every time a user loads a page with my php script, it sends me the html of the website, such as:

//the code goes here (get source code of the current website)
mail($my_email, " some subject", $SOURCE-OF-THIS-CURRENT-WEBSITE, "From: xxx");

I need this without using CURL! :)

Thank you.
LVL 2
sangeethaAsked:
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rdivilbissCommented:
$html = fopen($_SERVER['PHP_SELF']);

just a thought, not tested.
0
sangeethaAuthor Commented:
but upto my knowledge, $_SERVER variable takes the server information of where the file is located.

For example, if "http://www.test.com/parse.php" is the script which I need and if "http://www.example.com/file1.php" is the file that accesses the parse.php file, I need to know the source code of the "http://www.example.com/file1.php".
0
rdivilbissCommented:
That's what the fopen() is for.
0

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TeRReFCommented:
$html = file_get_contents($_SERVER['HTTP_REFERER']);
0
rdivilbissCommented:
file_get_contents(), a better choice.
0
JamesCsslCommented:
<?php
$handle = fopen($_SERVER['HTTP_REFERER'], "r");
$contents = '';
while (!feof($handle)) {
  $contents .= fread($handle, 8192);
}
fclose($handle);
?>

$contents contains the source of the referring site
0
rdivilbissCommented:
I suggested fopen() also, but really file_get_contents() is a better option for this particular question.
0
Computer101Commented:
Forced accept.

Computer101
EE Admin
0
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