“Notice: Undefined variable”, “Notice: Undefined i

Question

I'm running a PHP script and continue to receive errors like:

Notice: Undefined variable: my_variable_name in C:\wamp\www\mypath\index.php on line 10

Notice: Undefined index: my_index C:\wamp\www\mypath\index.php on line 11

Line 10 and 11 looks like this:

echo "My variable value is: " . $my_variable_name;
echo "My index value is: " . $my_array["my_index"];

What is the meaning of these error messages?

Why do they appear all of a sudden? I used to use this script for years and I've never had any problem.

How do I fix them?

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Answer 1

Generally because of "bad programming", and a possibility for mistakes now or later.

1. If it's a mistake, make a proper assignment to the variable first: $varname=0;
2. If it really is only defined sometimes, test for it: if (isset($varname)), before using it
3. If it's because you spelled it wrong, just correct that
4. Maybe even turn of the warnings in you PHP-settings

Answer 2

In reply to ""Why do they appear all of a sudden? I used to use this script for years and I've never had any problem."

It is very common for most sites to operate under the "default" error reporting of "Show all errors, but not 'notices' and 'deprecated'". This will be set in php.ini and apply to all sites on the server. This means that those "notices" used in the examples will be suppressed (hidden) while other errors, considered more critical, will be shown/recorded.

The other critical setting is the errors can be hidden (i.e. display_errors set to "off" or "syslog").

What will have happened in this case is that either the error_reporting was changed to also show notices (as per examples) and/or that the settings were changed to display_errors on screen (as opposed to suppressing them/logging them).

Why have they changed?

The obvious/simplest answer is that someone adjusted either of these settings in php.ini, or an upgraded version of PHP is now using a different php.ini from before. That's the first place to look.

However it is also possible to override these settings in

• .htconf (webserver configuration, including vhosts and sub-configurations)*
• .htaccess
• in php code itself

and any of these could also have been changed.

There is also the added complication that the web server configuration can enable/disable .htaccess directives, so if you have directives in .htaccess that suddenly start/stop working then you need to check for that.

(.htconf / .htaccess assume you're running as apache. If running command line this won't apply; if running IIS or other webserver then you'll need to check those configs accordingly)

Summary

• Check error_reporting and display_errors php directives in php.ini has not changed, or that you're not using a different php.ini from before.
• Check error_reporting and display_errors php directives in .htconf (or vhosts etc) have not changed
• Check error_reporting and display_errors php directives in .htaccess have not changed
• If you have directive in .htaccess, check if they are still permitted in the .htconf file
• Finally check your code; possibly an unrelated library; to see if error_reporting and display_errors php directives have been set there.

Answer 3

It means you are testing, evaluating, or printing a variable that you have not yet assigned anything to. It means you either have a typo, or you need to check that the variable was initialized to something first. Check your logic paths, it may be set in one path but not in another.

Answer 4

I use all time own useful function exst() which automatically declare variables.

Your code will be -

$greeting = "Hello, ".exst($user_name, 'Visitor')." from ".exst($user_location);


/** 
 * Function exst() - Checks if the variable has been set 
 * (copy/paste it in any place of your code)
 * 
 * If the variable is set and not empty returns the variable (no transformation)
 * If the variable is not set or empty, returns the $default value
 *
 * @param  mixed $var
 * @param  mixed $default
 * 
 * @return mixed 
 */

function exst( & $var, $default = "")
{
    $t = "";
    if ( !isset($var)  || !$var ) {
        if (isset($default) && $default != "") $t = $default;
    }
    else  {  
        $t = $var;
    }
    if (is_string($t)) $t = trim($t);
    return $t;
}

Answer 5

I didn't want to disable notice because it's helpful, but wanted to avoid too much typing.

My solution was this function:

function ifexists($varname)
{
  return(isset($$varname)?$varname:null);
}

So if I want to reference to $name and echo if exists, I simply write:

<?=ifexists('name')?>

For array elements:

function ifexistsidx($var,$index)
{
  return(isset($var[$index])?$var[$index]:null);
}

In page if I want to refer to $_REQUEST['name']:

<?=ifexistsidx($_REQUEST,'name')?>

Answer 6

the quick fix is to assign your variable to null at the top of your code

$user_location = null;