Undefined index name ошибка

Почему возникает ошибка

Ошибка undefined index появляется при попытке обращения к не существующему элементу массива:

<?php
$arr = [];
echo $arr['title'];

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

Notice: Undefined index: title in D:ProgramsOpenServerdomainstest.localindex.php on line 3

Как исправить ошибку

Если элемента в массиве нет, значит нужно ещё раз проверить логику программы и понять, почему в массиве нет тех данных, что вы ожидаете. Проверить, что по факту лежит в переменной можно с помощью функции var_dump():

$arr = [];
var_dump($arr);

При работе с массивами $_GET и $_POST нет гарантии, что клиент (браузер) отправил абсолютно все нужные нам данные. В этом случае можно добавить проверку на их существование:

<?php
if(!isset($_GET['body'], $_GET['title']))
	die('Пришли не все данные');

// Далее что-то делаем с данными

Если ключ массива существует не всегда, можно указать для него значение по-умолчанию:

<?php
if(isset($_GET['id']))
	$id = $_GET['id'];
else
	$id = 0;

Сокращённый синтаксис:

// С тернарным оператором
$id = isset($_GET['id']) ? $_GET['id'] : 0;

// С оператором объединения с null (PHP 7+)
$id = $_GET['id'] ?? 0;

Или если нужно сохранить значение по-умолчанию в сам массив:

<?php
if(!isset($arr['title']))
	$arr['title'] = '';

// Или короче (PHP 7+)
$arr['title'] = $arr['title'] ?? '';

// Или ещё короче (PHP 7.4+)
$arr['title'] ??= '';

Пишите в комментариях, если столкнулись с этой ошибкой и не можете найти решение.

Edit: See below for the solution

My composer.json:

{
    "name": "laravel/laravel",
    "type": "project",
    "description": "The Laravel Framework.",
    "keywords": [
        "framework",
        "laravel"
    ],
    "license": "MIT",
    "require": {
        "php": "^7.1.3",
        "area17/twill": "^2.1",
        "fideloper/proxy": "^4.0",
        "kalnoy/nestedset": "^5.0",
        "laravel/framework": "6.0.*",
        "laravel/helpers": "^1.2",
        "laravel/tinker": "^1.0"
    },
    "require-dev": {
        "barryvdh/laravel-debugbar": "3.*",
        "beyondcode/laravel-dump-server": "^1.0",
        "filp/whoops": "^2.0",
        "fzaninotto/faker": "^1.4",
        "mockery/mockery": "^1.0",
        "nunomaduro/collision": "^3.0",
        "phpunit/phpunit": "^7.5"
    },
    "config": {
        "optimize-autoloader": true,
        "preferred-install": "dist",
        "sort-packages": true
    },
    "extra": {
        "laravel": {
            "dont-discover": []
        }
    },
    "autoload": {
        "psr-4": {
            "App\": "app/"
        },
        "classmap": [
            "database/seeds",
            "database/factories"
        ]
    },
    "autoload-dev": {
        "psr-4": {
            "Tests\": "tests/"
        }
    },
    "minimum-stability": "dev",
    "prefer-stable": true,
    "scripts": {
        "post-autoload-dump": [
            "Illuminate\Foundation\ComposerScripts::postAutoloadDump",
            "@php artisan package:discover --ansi"
        ],
        "post-root-package-install": [
            "@php -r "file_exists('.env') || copy('.env.example', '.env');""
        ],
        "post-create-project-cmd": [
            "@php artisan key:generate --ansi"
        ]
    }
}

Output of composer diagnose:

Checking composer.json: OK
Checking platform settings: OK
Checking git settings: OK
Checking http connectivity to packagist: OK
Checking https connectivity to packagist: OK
Checking github.com rate limit: OK
Checking disk free space: OK
Checking pubkeys:
Tags Public Key Fingerprint: 57815BA2 7E54DC31 7ECC7CC5 573090D0 87719BA6 8F3BB723 4E5D42D0 84A14642
Dev Public Key Fingerprint: 4AC45767 E5EC2265 2F0C1167 CBBB8A2B 0C708369 153E328C AD90147D AFE50952
OK
Checking composer version: OK
Composer version: 2.0.1
PHP version: 7.3.7
PHP binary path: C:Program Filesphp7.3.7php.exe
OpenSSL version: OpenSSL 1.1.1c 28 May 2019
cURL version: 7.64.0 libz 1.2.11 ssl OpenSSL/1.1.1c
zip extension: OK

When I run this command:

composer update

I get the following output:

In PackageManifest.php line 122:

Undefined index: name

Script @php artisan package:discover —ansi handling the post-autoload-dump event returned with error code 1

The PHP undefined index notice signifies the usage of an unset index of a super global variable. Moreover, a similar notice will be generated when you call a variable without defining it earlier. As recurring notices disturb the flow of your program, this article contains all the causes and solutions of PHP notice: undefined index and similar notices in detail to get rid of them.

After reading this post, you’ll have enough alternative solutions to guard your program from such notices.

What Is Undefined Index in PHP?

The PHP undefined index is a notice that is generated as a result of accessing an unset index of a super global variable. Specifically, the notice is thrown while using the $_GET and $_POST superglobal variables. So, you might get the said notice while working with HTML forms.

What is the Cause of Undefined Index in PHP?

Surely, the stated variables store the data submitted through the forms based on the current form method. The process can be understood in a way that the input fields of the form are assigned unique names. Next, the values entered in the given input fields are accessed by passing the names of the stated fields as indices to the $_GET or $ _POST global variable.

Now, if no value is entered for a particular field and you try to access the same value, you’ll get the PHP notice: undefined index.

– Coding Example

The following example represents a script that throws an undefined index notice.

For instance: you have created a login form with a post method. The stated form consists of two fields as “username”, “password”, and a login button. Now, you would like to access the user credentials. Therefore, you’ll get the details entered by the user through the $_POST superglobal variable like $_POST[“username”] and $_POST[“password”].

But as the input fields will be empty for the first time, you will get two PHP undefined index notices below the form fields on loading the web page.

Here is a code snippet that depicts the above scenario in an understandable manner:

<!– creating a login form –>
<form action=”” method=”post”>
<input type=”text” name=”username” placeholder=”Enter Username”><br>
<input type=”text” name=”password” placeholder=”Enter Password”><br>
<input type=”submit” value=”Login” name=”submit”>
</form>
<?php
// accessing the values entered in the input fields
$username = $_POST[“username”];
$password = $_POST[“password”];
?>

How to Solve the Error?

Undeniably, the PHP undefined index notices printed below your form fields pose a bad impact on your users while decreasing the user experience. Therefore, the mentioned notices need to be removed as soon as possible. So, here you’ll implement the isset() function to avoid the PHP notice: undefined index on loading the web page.

The isset() function accepts a variable to check if the value of the same variable is not null. Also, you can check the nullability of more than one variable at the same time.

Here is the syntax for your reference: isset(variable, …).

– Coding Example of Eliminating the Warning

For example, you are working on a contact form that consists of name, email, and feedback fields. Moreover, you have a submit button below the fields. Similar to the above example, here you’ll get the PHP undefined index notices on loading the given web page for the first time. So, you’ll use the isset() function to check if the form has been submitted before accessing the values of the input fields.

Eventually, you’ll make the notices go away as seen in this code block:

<!– creating a contact form –>
<form action=”” method=”post”>
<input type=”text” name=”name” placeholder=”Enter Your Name”><br>
<input type=”text” name=”email” placeholder=”Enter Your Email”><br>
<textarea name=”feedback” cols=”30″ rows=”10″></textarea><br>
<input type=”submit” value=”Submit” name=”submitBtn”>
</form>
<?php
// accessing the values entered in the input fields on form submission
if(isset($_POST[“submitBtn”])){
$name = $_POST[“name”];
$email = $_POST[“email”];
$feedback = $_POST[“feedback”];
}
?>

PHP Undefined Variable Notices

Another kind of notice that somehow resembles the PHP notice: undefined index is the PHP undefined variable. You will see such notice on your screen when you try to use a variable without defining it earlier in your program. Hence, the reason behind getting the PHP undefined variable notice can be either a human error, an uncertainty, or a situation where you haven’t decided on the value of the variable yet. Well, whatever is the reason, all of the given situations and their possible solutions have been stated below for your convenience:

– Solving the Error Caused by Human Error

So, if you’ve forgotten to define the given variable then define it before using the same for avoiding the PHP undefined variable notice. Also, a misspelled variable name can result in throwing the same notice. Therefore, check out the spellings and the variable definitions before running your script to have a notice-free script execution.

– Solving the Error Caused Due To Uncertainty

Suppose you aren’t sure if the variable has been defined already or not. Plus, you don’t want to redefine the given variable then you can use the isset() function to see the current status of the variable and deal with it accordingly like this:

<?php
// creating an array
$array1 = array();
// checking if the $color variable has been defined already
if((isset($color))){
// adding the variable in the array
$array1[] .= $color;
// printing the array
print_r($array1);
}
else
{
// printing a friendly statement to define the color variable
echo “Please define the color variable.”;
}
?>

– Solving the Error Caused Due To Undecided Value

Would you like to proceed with using a particular variable in your program instead of thinking about a perfect value for the same? If this is the case then begin with defining the variable with an empty string to avoid the PHP undefined variable notice as seen in the code block here:

<?php
// creating an empty variable
$name = “”;
// using the variable in a statement
echo “The name of the girl was $name and she loved to create websites.”;
// output: The name of the girl was and she loved to create websites.
?>

Undefined Offset in PHP

Are you currently working with arrays and getting an undefined offset notice? Well, it would be helpful to inform you that the said notice will appear on your screen if you call an undefined index or named key. Therefore, it is the way of PHP to tell you that you are trying to access a key that does not exist in the given array.

– Code Block Depicting the Cause of Undefined Offset Notice

For instance, you have an array of different cars in which the brands of the cars are set as keys while the colors of the cars are set as values. As you created the mentioned array some days back, now you mistakenly called a car brand that doesn’t even exist. Consequently, you’ll see PHP coming with a notice of undefined offset similar to the results of the below code block:

<?php
// creating an array of cars
$cars = array(
“Audi” => “Blue”,
“Ford” => “Black”,
“Toyota” => “Red”,
“Bentley” => “Grey”,
“BMW” => “White”
);
// accessing a car brand that doesn’t exist
echo $cars[“Mercedes-Benz”];
?>

– Main Solutions for Undefined Offset

Interestingly, there are two ways that will help in avoiding the PHP undefined offset notice including the isset() function and the array_key_exists() function. Indeed, you can call any of the given functions to check if the key that you are planning to access already exists in your given array.

However, the checking procedure carried out by the isset() differs from the one implemented by the array_key_exists() function. Still, you are free to use any of the stated functions to avoid the undefined offset notice.

Continuing with the same example of the cars array. Here, you’ll use either the isset() or the array_key_exists() function to check if a particular car brand exists in the “cars” array. Next, you’ll use the stated car brand in your program based on the result returned by the said functions.

Please see this code snippet for effective implementation of the above functions:

<?php
// using the isset() function
if(isset($cars[“Mercedes-Benz”])){
echo $cars[“Mercedes-Benz”];
}
// or use the array_key_exists() function
if(array_key_exists(“Mercedes-Benz”,$cars)){
echo $cars[“Mercedes-Benz”];
}
?>

Conclusion

The PHP undefined index and similar notices arise because of using unset values. But thankfully, the given notices can be eliminated by carefully using the variables and enabling the extra checking system. Also, here is a list of the points that will help you in staying away from such disturbing notices:

• The PHP undefined index notice will be generated as a result of using the unset index of the $_POST or $_GET superglobal variable
• You can get rid of the PHP undefined index notice by using the isset() function
• The PHP undefined variable notice can be removed by either using the isset() function or setting the variable value as an empty string
• An undefined offset notice will be generated when you call an array key that doesn’t exist
• You can use the isset() or the array_key_exists() function to access the array keys without getting any notices

Although the notices don’t stop your program’s execution, having the same on your screen can be annoying leading you to look for solutions like the ones shared above.

While working in PHP, you will come across two methods called $_POST and $_GET. These methods are used for obtaining values from the user through a form. When using them, you might encounter an error called “Notice: Undefined Index”. 

This error means that within your code, there is a variable or constant that has no value assigned to it. But you may be trying to use the values obtained through the user form in your PHP code.

The error can be avoided by using the isset() function. This function will check whether the index variables are assigned a value or not, before using them.

Undefined Index PHP Error

An undefined index is a ‘notice’ such as the following:

“Notice: Undefined variable,” 

“Notice: Undefined index” and “Notice: Undefined offset.”

As you can see above are all notices, here are two ways to deal with such notices.

1) Ignore such notices
2) Resolve such notices.

How to Ignore PHP Notice: Undefined Index

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

1. php.ini

Open php.ini file in your favourite 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. 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.’

Solution or Fix for PHP Notice: Undefined Index

Cause of error:

This error occurs with $ _POST and $ _GET method when you use index or variables which have not set through $ _POST or $ _GET method, but you are already using their value in your PHP code.

Undefined index in PHP $_get

Example using $_GET

In the following example, we have used two variables ‘ names’ & ‘age,’ but we did set only the name variable through the $_GET method, that’s why it throws the notice.

http://yoursite.com/index.php?name=ram

<?php 
$name = $_GET['name'];
$age = $_GET['age'];

echo $name;
echo $age;
?>

OUTPUT:

Notice: Undefined index: age index.php on line 5

Solution

To solve such error, you can use the isset() function, which will check whether the index or variable is set or not, If not then don’t use it.

if(isset($_GET[index error name]))

Code with Error resolved using isset() function:

http://yoursite.com/index.php?name=ram

<?php
if(isset($_GET['name'])){
      $name = $_GET['name']; 
 }else{
      $name = "Name not set in GET Method";
 }
if(isset($_GET['age'])){
      $name = $_GET['age']; 
 }else{
      $name = "<br>Age not set in GET Method";
 }
echo $name;
echo $age;
?>

OUTPUT:

ram
Age not set in GET Method

Set Index as blank

We can also set the index as blank index:

// example with $_POST method

$name = isset($_POST['name']) ? $_POST['name'] : '';
$name = isset($_POST['age']) ? $_POST['age'] : '';

// example with $_GET method

$name = isset($_GET['name']) ? $_GET['name'] : '';
$name = isset($_GET['age']) ? $_GET['age'] : '';

Notice: Undefined Variable

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

Example:

<?php 
$name='RAM';

echo $name;
echo $age;
?>

Output:

Notice: Undefined variable: age in D:xampphtdocstestsite.locindex.php on line 7

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

Solutions:

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

<?php 
global $name;
global $age; 
echo $name;
?>

<?php
if(isset($name)){echo $name;}
if(isset($age)){echo $age;}
?>

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

Notice: Undefined Offset

This type of error occurs with arrays when we use the key of an array, which is not set.

In the following, given an example, we are displaying the value store in array key 1, but we did not set while declaring array “$colorarray.”

Example:

<?php 
// declare an array with key 2, 3, 4, 5 
$colorarray = array(2=>'Red',3=>'Green',4=>'Blue',5=>'Yellow');

// echo value of array at offset 1.
echo $colorarray[1];
?>

Output: 

Notice: Undefined offset: 1 in index.php on line 5

Solutions:

Check the value of offset array with function isset() & empty(), and use array_key_exists() function to check if key exist or not.

<?php 
$colorarray = array(2=>'Red',3=>'Green',4=>'Blue',5=>'Yellow');

// isset() function to check value at offset 1 of array
if(isset($colorarray[1])){echo $colorarray[1];}

// empty() function to check value at offset 1 of array
if(!empty($colorarray[1])){echo $colorarray[1];}

// array_key_exists() of check if key 1 is exist or not
echo array_key_exists(1, $colorarray);
?>