Php handle undefined variable

How to handle undefined variable in php

Solution 1: isset Solution 2: If you want to check if a parameter is passed in through the URL, you can use isset() http://php.net/manual/en/function.isset.php This will check to see if the parameter exists Solution 3: You can use ternary operator Solution 1: Strictly speaking, all you need is to change to A couple things to note here: URL parameters become items in the array, not independent variables. So, that all in mind, a much better implementation for index.php would be: Solution 2: You need to define before calling it, you could try: Of course you need to have a default page in case is not set in the URL

PHP catching undefined variable error [duplicate]

undefined variables can be corrected by DEFINING them.

defines the variable. you will not get any error or warning.

Why are you using try-catch for that? Just use isset

just use isset() to determine if the variable is set.

Undefined index in php Code Example, Your array index has no value or is not referencing anything, // The easiest way to overcome this is to simply check whether // it has been defined if

PHP: handle undefined exception

If you want to check if a parameter is passed in through the URL, you can use isset()

This will check to see if the parameter exists

// This will evaluate to TRUE so the text will be printed. if (isset($var))

You can use ternary operator

MySQL query handle undefined, I have the following code, which works fine. Queries are executed, everything is OK. BUT I am getting a error from PHP: Notice: Undefined

PHP include / undefined variable

Strictly speaking, all you need is to change

A couple things to note here:

• URL parameters become items in the $_GET array, not independent variables.
• include doesn’t need parenthesis. It’s not wrong to include them, but it’s not necessary. See the documentation.
• You shouldn’t actually use this implementation as it is because it’s extremely insecure. A very simply bit of URL manipulation, and some fellow with poor intentions could figure out a lot about your file structure. At the very least, you should check for a particular extension, or append a particular directory name in the include statement.
• You also shouldn’t use this implementation as it is, because it doesn’t deal with non-existent files gracefully. If the requested content file doesn’t exist, you should send an error message and a 404 header. Sending the header has to be done before any of the content is sent, so a check for whether files exist should be done early in the process.

So, that all in mind, a much better implementation for index.php would be:

You need to define $v before calling it, you could try:

Of course you need to have a default page in case v is not set in the URL hence the else home.php

To pass variables in the URL, you need to use the $_GET array. It will be populated automatically before your script is ran, so you only need to read from it. For example, $_GET[‘v’] will contain the value you have passed in though the URL ( ?v=. ).

Note that it is always best to first check if the value is set, or else it will throw a nasty error at you in case it isn’t. The following code is a simple way to handle an unset parameter:

PHP: handle undefined exception, On having error such as undefined index, undefined variable etc.. will be handle by handleError function. public function handleError($code

MySQL query handle undefined

Using dollar sign for class property is unnecessary, You can change $this->$handle to $this->handle

function __construct($hostname, $user, $pass, $databasename) < $this->handle = mysqli_connect($hostname, $user, $pass); if(!$this->handle) < $this->seterror(); return false; > $handle = mysqli_select_db($this->handle, $databasename); if(!$handle) < $this->seterror(); return false; > return true; > function query($text) < $result = mysqli_query($this->handle,$text); // LINE 43 if(!$result) < $this->seterror(); $this->geterror(); // temp return false; > return $result; > 

handle should be a property of your class declared with «public», «private» or «protected» keyword. You should not use «var» to declare property specially if you are using recent version of PHP.

Like it was said before. If you declare the property properly, you will have to change

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Исправление ошибки «Notice: undefined variable» в PHP

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

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:\Programs\OpenServer\domains\test.local\index.php on line 2

Как исправить ошибку

Нужно объявить переменную перед обращением к ней:

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

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

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. 🙂

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Notice: Undefined Variable in PHP

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:

Output:

STechies Notice: Undefined variable: age in \testsite.loc\varaible.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.

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:

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

Set Index as blank

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:

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.

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

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[Solved]: Notice: Undefined variable in PHP

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

Example 1

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

Example 2

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

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

 else < $result = "The name has not been set"; >echo $result; //Output: The name is Raju Rastogi echo "
" //Example 2 if(isset($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.

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.

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

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

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 & ~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.

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