“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

Try these

Q1: this notice means $varname is not defined at current scope of the script.

Q2: Use of isset(), empty() conditions before using any suspicious variable works well.

// recommended solution for recent PHP versions
$user_name = $_SESSION['user_name'] ?? '';

// pre-7 PHP versions
$user_name = '';
if (!empty($_SESSION['user_name'])) {
     $user_name = $_SESSION['user_name'];
}

Or, as a quick and dirty solution:

// not the best solution, but works
// in your php setting use, it helps hiding site wide notices
error_reporting(E_ALL ^ E_NOTICE);

---

Note about sessions:

• When using sessions, session_start(); is required to be placed inside all files using sessions.

• http://php.net/manual/en/features.sessions.php

Answer 2

Using a ternary is simple, readable, and clean:

Pre PHP 7
Assign a variable to the value of another variable if it's set, else assign null (or whatever default value you need):

$newVariable = isset($thePotentialData) ? $thePotentialData : null;

PHP 7+
The same except using Null Coalescing Operator. There's no longer a need to call isset() as this is built in, and no need to provide the variable to return as it's assumed to return the value of the variable being checked:

$newVariable = $thePotentialData ?? null;

Both will stop the Notices from the OP question, and both are the exact equivalent of:

if (isset($thePotentialData)) {
    $newVariable = $thePotentialData;
} else {
    $newVariable = null;
}

 
If you don't require setting a new variable then you can directly use the ternary's returned value, such as with echo, function arguments, etc:

Echo:

echo 'Your name is: ' . isset($name) ? $name : 'You did not provide one';

Function:

$foreName = getForeName(isset($userId) ? $userId : null);

function getForeName($userId)
{
    if ($userId === null) {
        // Etc
    }
}

The above will work just the same with arrays, including sessions etc, replacing the variable being checked with e.g.:
$_SESSION['checkMe']
or however many levels deep you need, e.g.:
$clients['personal']['address']['postcode']

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Suppression:

It is possible to suppress the PHP Notices with @ or reduce your error reporting level, but it does not fix the problem, it simply stops it being reported in the error log. This means that your code still tried to use a variable that was not set, which may or may not mean something doesn't work as intended - depending on how crucial the missing value is.

You should really be checking for this issue and handling it appropriately, either serving a different message, or even just returning a null value for everything else to identify the precise state.

If you just care about the Notice not being in the error log, then as an option you could simply ignore the error log.

Answer 3

Probably you were using old PHP version until and now upgraded PHP thats the reason it was working without any error till now from years. until PHP4 there was no error if you are using variable without defining it but as of PHP5 onwards it throws errors for codes like mentioned in question.

Answer 4

I asked a question about this and I was referred to this post with the message:

This question already has an answer here:

“Notice: Undefined variable”, “Notice: Undefined index”, and “Notice: Undefined offset” using PHP

I am sharing my question and solution here:

This is the error:

Line 154 is the problem. This is what I have in line 154:

153    foreach($cities as $key => $city){
154        if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){

I think the problem is that I am writing if conditions for the variable $city, which is not the key but the value in $key => $city. First, could you confirm if that is the cause of the warning? Second, if that is the problem, why is it that I cannot write a condition based on the value? Does it have to be with the key that I need to write the condition?

UPDATE 1: The problem is that when executing $citiesCounterArray[$key], sometimes the $key corresponds to a key that does not exist in the $citiesCounterArray array, but that is not always the case based on the data of my loop. What I need is to set a condition so that if $key exists in the array, then run the code, otherwise, skip it.

UPDATE 2: This is how I fixed it by using array_key_exists():

foreach($cities as $key => $city){
    if(array_key_exists($key, $citiesCounterArray)){
        if(($city != 'London') && ($city != 'Madrid') && ($citiesCounterArray[$key] >= 1)){

Answer 5

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 6

why not keep things simple?

<?php
error_reporting(E_ALL); // making sure all notices are on

function idxVal(&$var, $default = null) {
         return empty($var) ? $var = $default : $var;
  }

echo idxVal($arr['test']);         // returns null without any notice
echo idxVal($arr['hey ho'], 'yo'); // returns yo and assigns it to array index, nice

?>