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Ajax request using PHP forwarding

<?php  
  header('Content-Type: application/xml');  
  $tmpFile = 'tmpFile.txt';  
  $val = $_GET["http://rss.news.yahoo.com/rss/us.rss"]; 
  echo ($val); 
  $curlHandle = curl_init($val);  
  $filePointer = fopen($tmpFile, "w");  
  curl_setopt($curlHandle, CURLOPT_FILE, $filePointer);  
  curl_exec($curlHandle);  
  curl_close($curlHandle);  
  fclose($filePointer);  
  $linesArr = file($tmpFile);  
  foreach($linesArr as $eachLine){  
    echo($eachLine);  
  }  
?>

I don't know why this code is not working. Can someone help me

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otengkwaku
Junior Poster in Training
92 posts since Mar 2012
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What's this:

 $val = $_GET["http://rss.news.yahoo.com/rss/us.rss"];

Do you have a querystring with a parameter of that name? Sounds unlikely.

diafol
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10,636 posts since Oct 2006
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Solved Threads: 1,508
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sorry for that it is realy $val = $_GET["fwd"]

the posted code is to get the content of a web site and save it in a file

otengkwaku
Junior Poster in Training
92 posts since Mar 2012
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I followed the link and it doesn't seem to exist.

diafol
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but should the code on the whole work?
when i run the code with a deffrent url it gives me the error

<b>Notice</b>:  Undefined index: https://www.google.com in <b>C:\Users\spirit\Documents\xampp\htdocs\News-Reader-Code\news-web-code\bridge2.php</b> on line <b>4</b><br />
otengkwaku
Junior Poster in Training
92 posts since Mar 2012
Reputation Points: 23
Solved Threads: 1
Skill Endorsements: 8

sounds like you are still using the weird key:

$val = $_GET["http://www.google.com"]; 

?

diafol
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Question Answered as of 3 Months Ago by diafol

This question has already been solved: Start a new discussion instead

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