Report undefined variable как исправить

Почему возникает ошибка

Ошибка undefined variable появляется при попытке обратиться к не существующей (не объявленной ранее) переменной:

<?php
echo $text;

Если в настройках PHP включено отображение ошибок уровня E_NOTICE, то при запуске этого кода в браузер выведется ошибка:

Notice: Undefined variable: text in D:ProgramsOpenServerdomainstest.localindex.php on line 2

Как исправить ошибку

Нужно объявить переменную перед обращением к ней:

$a = '';
echo $a;

Нет уверенности, что переменная будет существовать? Можно указать значение по-умолчанию:

<?php
if(!isset($text))
    $text = '';

echo $text;

Или сокращённые варианты:

<?php
// С оператором объединения с null (PHP 7+)
$text = $text ?? '';

// С оператором присваивания значения null (PHP 7.4+)
$text ??= '';

Есть ещё один вариант исправления этой ошибки – отключить отображение ошибок уровня E_NOTICE:

<?php
error_reporting(E_ALL & ~E_NOTICE);
echo $a; // Ошибки не будет

Не рекомендую этот вариант. Скрытие ошибок вместо их исправления – не совсем правильный подход.

Кроме этого, начиная с PHP 8 ошибка undefined variable перестанет относиться к E_NOTICEи так легко отключить её уже не удастся.

Если ошибка появилась при смене хостинга

Часто ошибка возникает при переезде с одного сервера на другой. Практически всегда причина связана с разными настройками отображения ошибок на серверах.

По-умолчанию PHP не отображает ошибки уровня E_Notice, но многие хостинг-провайдеры предпочитают настраивать более строгий контроль ошибок. Т.е. на старом сервере ошибки уже были, но игнорировались сервером, а новый сервер таких вольностей не допускает.

Остались вопросы? Добро пожаловать в комментарии. 🙂

  John Mwaniki /   10 Dec 2021

This error, as it suggests, occurs when you try to use a variable that has not been defined in PHP.

Example 1

<?php
echo $name;
?>

Output

Notice: Undefined variable: name in /path/to/file/file.php on line 2.

Example 2

<?php
$num1 = 27;
$answer = $num1 + $num2;
echo $answer;
?>

Output

Notice: Undefined variable: num2 in /path/to/file/file.php on line 3.

In our above two examples, we have used a total of 4 variables which include $name, $num1, $num2, and $answer.

But out of all, only two ($name and $num2) have resulted in the “undefined variable” error. This is because we are trying to use them before defining them (ie. assigning values to them).

In example 1, we are trying to print/display the value of the variable $name, but we had not yet assigned any value to it.

In example 2, we are trying to add the value of $num1 to the value of $num2 and assign their sum to $answer. However, we have not set any value for $num2.

To check whether a variable has been set (ie. assigned a value), we use the in-built isset() PHP function.

Syntax

isset($variable)

We pass the variable name as the only argument to the function, where it returns true if the variable has been set, or false if the variable has not been set.

Example

<?php
//Example 1
$name = "Raju Rastogi";
if(isset($name)){
  $result = "The name is $name";
}
else{
  $result = "The name has not been set";
}
echo $result;
//Output: The name is Raju Rastogi
echo "<br>"


//Example 2
if(isset($profession)){
  $result = "My profession is $profession";
}
else{
  $result = "The profession has not been set";
}
echo $result;
//Output: The profession has not been set
?>

In our first example above, we have defined a variable ($name) by creating it and assigning it a value (Raju Rastogi) and thus the isset() function returns true.

In our second example, we had not defined our variable ($profession) before passing to the isset() function and thus the response is false.

The Fix for Undefined variable error

Here are some ways in which you can get rid of this error in your PHP program.

1. Define your variables before using them

Since the error originates from using a variable that you have not defined (assigned a value to), the best solution is to assign a value to the variable before using it anywhere in your program.

For instance, in our first example, the solution is to assign a value to the variable $name before printing it.

<?php
$name = "Farhan Qureshi";
echo $name;
//Output: Farhan Qureshi
?>

The above code works without any error.

2. Validating variables with isset() function

Another way to go about this is to validate whether the variables have been set before using them.

<?php
$num1 = 27;
if(isset($num1) && isset($num2)){
$answer = $num1 + $num2;
echo $answer;
}
?>

The addition operation will not take place because one of the required variables ($num2) has not been set.

<?php
$num1 = 27;
$num2 = 8;
if(isset($num1) && isset($num2)){
$answer = $num1 + $num2;
echo $answer;
}
//Oputput: 35
?>

The addition this time will take place because the two variables required have been set.

3. Setting the undefined variable to a default value

You can also check whether the variables have been defined using the isset() function and if not, assign them a default value.

For instance, you can set a blank “” value for variables expected to hold a string value, and a 0 for those values expect to hold a numeric value.

Example

<?php
$name = "John Doe";
$name = isset($name) ? $name : '';
$age= isset($age) ? $age: 0;
echo "My name is $name and I am $age yrs old.";
//Output: My name is John Doe and I am 0 yrs old.
?>

4. Disabling Notice error reporting

This is always a good practice to hide errors from the website visitor. In case these errors are visible on the website to the web viewers, then this solution will be very useful in hiding these errors.

This option does not prevent the error from happening, it just hides it.

Open the php.ini file in a text editor, and find the line below:

error_reporting = E_ALL

Replace it with:

error_reporting = E_ALL & ~E_NOTICE

Now the ‘NOTICE’ type of errors won’t be shown again. However, the other types of errors will be shown.

Another way of disabling these errors is to add the line of code below on the top of your PHP code.

error_reporting (E_ALL ^ E_NOTICE);

That’s all for this article. It’s my hope that it has helped you solve the error.

Notice: Undefined variable

This error means that within your code, there is a variable or constant which is not set. But you may be trying to use that variable.

The error can be avoided by using the isset() function.This function will check whether the variable is set or not.

Error Example:

<?php
$name='Stechies';
echo $name;
echo $age;
?>

Output:

STechies
Notice: Undefined variable: age in testsite.locvaraible.php on line 4

In the above example, we are displaying value stored in the ‘name’ and ‘age’ variable, but we didn’t set the ‘age’ variable.

Here are two ways to deal with such notices.

1. Resolve such notices.
2. Ignore such notices.

Fix Notice: Undefined Variable by using isset() Function

This notice occurs when you use any variable in your PHP code, which is not set.

Solutions:

To fix this type of error, you can define the variable as global and use the isset() function to check if the variable is set or not.

Example:

<?php
global $name;
global $age;
$name = 'Stechies';

if(!isset($name)){
$name = 'Variable name is not set';
}
if(!isset($age)){
$age = 'Varaible age is not set';
}
echo 'Name: ' . $name.'<br>';
echo 'Age: ' . $age;
?>

Set Index as blank

<?php
$name = 'Stechies';

// Set Variable as Blank
$name = isset($name) ? $name : '';
$age= isset($age) ? $age: '';

echo 'Name: ' . $name.'<br>';
echo 'Age: ' . $age;
?>

Ignore PHP Notice: Undefined variable

You can ignore this notice by disabling reporting of notice with option error_reporting.

1. Disable Display Notice in php.ini file

Open php.ini file in your favorite editor and search for text “error_reporting” the default value is E_ALL. You can change it to E_ALL & ~E_NOTICE.

By default:

error_reporting = E_ALL

Change it to:

error_reporting = E_ALL & ~E_NOTICE

Now your PHP compiler will show all errors except ‘Notice.’

2. Disable Display Notice in PHP Code

If you don’t have access to make changes in the php.ini file, In this case, you need to disable the notice by adding the following code on the top of your PHP page.

<?php error_reporting (E_ALL ^ E_NOTICE); ?>

Now your PHP compiler will show all errors except ‘Notice.’