Check if a variable is undefined in PHP

You can use -

$isTouch = isset($variable);

It will return true if the $variable is defined. If the variable is not defined it will return false.

Note: It returns TRUE if the variable exists and has a value other than NULL, FALSE otherwise.

If you want to check for false, 0, etc., you can then use empty() -

$isTouch = empty($variable);

empty() works for -

• "" (an empty string)
• 0 (0 as an integer)
• 0.0 (0 as a float)
• "0" (0 as a string)
• NULL
• FALSE
• array() (an empty array)
• $var; (a variable declared, but without a value)

Another way is simply:

if($test){
    echo "Yes 1";
}
if(!is_null($test)){
    echo "Yes 2";
}

$test = "hello";

if($test){
    echo "Yes 3";
}

Will return:

"Yes 3"

The best way is to use isset(). Otherwise you can have an error like "undefined $test".

You can do it like this:

if(isset($test) && ($test!==null))

You'll not have any error, because the first condition isn't accepted.

The isset() function does not check if a variable is defined.

It seems you've specifically stated that you're not looking for isset() in the question. I don't know why there are so many answers stating that isset() is the way to go, or why the accepted answer states that as well.

It's important to realize in programming that null is something. I don't know why it was decided that isset() would return false if the value is null.

To check if a variable is undefined you will have to check if the variable is in the list of defined variables, using get_defined_vars(). There is no equivalent to JavaScript's undefined (which is what was shown in the question, no jQuery being used there).

In the following example it will work the same way as JavaScript's undefined check.

$isset = isset($variable);
var_dump($isset); // false

But in this example, it won't work like JavaScript's undefined check.

$variable = null;
$isset = isset($variable);
var_dump($isset); // false

$variable is being defined as null, but the isset() call still fails.

So how do you actually check if a variable is defined? You check the defined variables.

Using get_defined_vars() will return an associative array with keys as variable names and values as the variable values. We still can't use isset(get_defined_vars()['variable']) here because the key could exist and the value still be null, so we have to use array_key_exists('variable', get_defined_vars()).

$variable = null;
$isset = array_key_exists('variable', get_defined_vars());
var_dump($isset); // true


$isset = array_key_exists('otherVariable', get_defined_vars());
var_dump($isset); // false

However, if you're finding that in your code you have to check for whether a variable has been defined or not, then you're likely doing something wrong. This is my personal belief as to why the core PHP developers left isset() to return false when something is null.

To check if a variable is set you need to use the isset function.

$lorem = 'potato';

if(isset($lorem)){
    echo 'isset true' . '<br />';
}else{
    echo 'isset false' . '<br />';
}

if(isset($ipsum)){
    echo 'isset true' . '<br />';
}else{
    echo 'isset false' . '<br />';
}

This code will print:

isset true
isset false

Read more in isset.

You can use the ternary operator to check whether the value is set by POST/GET or not. Something like this:

$value1 = $_POST['value1'] = isset($_POST['value1']) ? $_POST['value1'] : '';
$value2 = $_POST['value2'] = isset($_POST['value2']) ? $_POST['value2'] : '';
$value3 = $_POST['value3'] = isset($_POST['value3']) ? $_POST['value3'] : '';
$value4 = $_POST['value4'] = isset($_POST['value4']) ? $_POST['value4'] : '';