Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3

StackOverflow https://stackoverflow.com/questions/16684214

  •  30-05-2022
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Question

I am having a problem in the shortcut links from this site: http://smileshort.com/short-anonymous-links-api.html

Use api: http://smileshort.com/api.php?key=534287562&url=google.com

Show me this problem

Warning: Unknown: Your script possibly relies on a session side-effect which existed until PHP 4.2.3. Please be advised that the session extension does not consider global variables as a source of data, unless register_globals is enabled. You can disable this functionality and this warning by setting session.bug_compat_42 or session.bug_compat_warn to off, respectively in Unknown on line 0

when i Use this function

<?php

function get_vgd($url)
{
$apiurl = "http://smileshort.com/api.php?key=890479270&url=$url";
$ch = curl_init();
$timeout = 3;
curl_setopt($ch,CURLOPT_URL,$apiurl);
curl_setopt($ch,CURLOPT_RETURNTRANSFER,1);
curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout);
$data = curl_exec($ch);
curl_close($ch);
return $data;
}
echo get_vgd("http://www.google.com");

?>
Was it helpful?

Solution

There is a session variable given the same name as some existing global variable (i.e. both $_SESSION['name'] and $name exists).
Rename either of them.

It should be a quite duplicate question, as it just occurred to me.

OTHER TIPS

There can be another reason for this error. In case somebody else comes to this thread and knows that there is no duplicate variables.

If you do something like

unset($var);
$_SESSION['sessVar'] = $var;

you will get the same warning because you set an undefined variable to a session variable. The unset is usually not there, it was just for illustration purposes :)

I've just had this problem, and the reason for problem was

a variable called on the page was not defined.

It put me off track and got me quite concerned but now I've defined the $Var it no longer prints the Warning.

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