Ajax request using PHP forwarding

Question

otengkwaku 30 Newbie Poster

11 Years Ago

<?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

javascript php

2 Contributors
5 Replies
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3 Hours Discussion Span
Latest Post 11 Years Ago Latest Post by diafol

diafol

11 Years Ago

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

11 Years Ago

I followed the link and it doesn't seem to exist.

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otengkwaku 30 Newbie Poster · Answer 1 · 2013-01-25T22:36:24+00:00

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 30 Newbie Poster · Answer 2 · 2013-01-25T23:34:30+00:00

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 />

diafol · Answer 3 · 2013-01-26T01:42:57+00:00

sounds like you are still using the weird key:

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

?