In where shall I use isset() and !empty()

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I read somewhere that the isset() function treats an empty string as TRUE, therefore isset() is not an effective way to validate text inputs and text boxes from a HTML form.

So you can use empty() to check that a user typed something.

  1. Is it true that the isset() function treats an empty string as TRUE?

  2. Then in which situations should I use isset()? Should I always use !empty() to check if there is something?

For example instead of

if(isset($_GET['gender']))...

Using this

if(!empty($_GET['gender']))...
19 Answers

isset vs. !empty

FTA:

"isset() checks if a variable has a value including (False, 0 or empty string), but not NULL. Returns TRUE if var exists; FALSE otherwise.

On the other hand the empty() function checks if the variable has an empty value empty string, 0, NULL or False. Returns FALSE if var has a non-empty and non-zero value."

In the most general way :

  • isset tests if a variable (or an element of an array, or a property of an object) exists (and is not null)
  • empty tests if a variable (...) contains some non-empty data.


To answer question 1 :

$str = '';
var_dump(isset($str));

gives

boolean true

Because the variable $str exists.


And question 2 :

You should use isset to determine whether a variable exists ; for instance, if you are getting some data as an array, you might need to check if a key isset in that array.
Think about $_GET / $_POST, for instance.

Now, to work on its value, when you know there is such a value : that is the job of empty.

isset is intended to be used only for variables and not just values, so isset("foobar") will raise an error. As of PHP 5.5, empty supports both variables and expressions.

So your first question should rather be if isset returns true for a variable that holds an empty string. And the answer is:

$var = "";
var_dump(isset($var));

The type comparison tables in PHP’s manual is quite handy for such questions.

isset basically checks if a variable has any value other than null since non-existing variables have always the value null. empty is kind of the counter part to isset but does also treat the integer value 0 and the string value "0" as empty. (Again, take a look at the type comparison tables.)

isset() is not an effective way to validate text inputs and text boxes from a HTML form

You can rewrite that as "isset() is not a way to validate input." To validate input, use PHP's filter extension. filter_has_var() will tell you whether the variable exists while filter_input() will actually filter and/or sanitize the input.

Note that you don't have to use filter_has_var() prior to filter_input() and if you ask for a variable that is not set, filter_input() will simply return null.

isset() is used to check if the variable is set with the value or not and Empty() is used to check if a given variable is empty or not.

isset() returns true when the variable is not null whereas Empty() returns true if the variable is an empty string.

isset($variable) === (@$variable !== null)
empty($variable) === (@$variable == false)

I came here looking for a quick way to check if a variable has any content in it. None of the answers here provided a full solution, so here it is:


It's enough to check if the input is '' or null, because:

Request URL .../test.php?var= results in $_GET['var'] = ''

Request URL .../test.php results in $_GET['var'] = null


isset() returns false only when the variable exists and is not set to null, so if you use it you'll get true for empty strings ('').

empty() considers both null and '' empty, but it also considers '0' empty, which is a problem in some use cases.

If you want to treat '0' as empty, then use empty(). Otherwise use the following check:

$var .'' !== '' evaluates to false only for the following inputs:

  • ''
  • null
  • false

I use the following check to also filter out strings with only spaces and line breaks:

function hasContent($var){
    return trim($var .'') !== '';
}

Using empty is enough:

if(!empty($variable)){
    // Do stuff
}

Additionally, if you want an integer value it might also be worth checking that intval($variable) !== FALSE.

    $var = '';
// Evaluates to true because $var is empty
if ( empty($var) ) {
echo '$var is either 0, empty, or not set at all';
}
// Evaluates as true because $var is set
if ( isset($var) ) {
 echo '$var is set even though it is empty';
    }

Source: Php.net

!empty will do the trick. if you need only to check data exists or not then use isset other empty can handle other validations

<?php
$array = [ "name_new" => "print me"];

if (!empty($array['name'])){
   echo $array['name'];
}

//output : {nothing}

////////////////////////////////////////////////////////////////////

$array2 = [ "name" => NULL];

if (!empty($array2['name'])){
   echo $array2['name'];
}

//output : {nothing}

////////////////////////////////////////////////////////////////////

$array3 = [ "name" => ""];

if (!empty($array3['name'])){
   echo $array3['name'];
}

//output : {nothing}  

////////////////////////////////////////////////////////////////////

$array4 = [1,2];

if (!empty($array4['name'])){
   echo $array4['name'];
}

//output : {nothing}

////////////////////////////////////////////////////////////////////

$array5 = [];

if (!empty($array5['name'])){
   echo $array5['name'];
}

//output : {nothing}

?>

When in doubt, use this one to check your Value and to clear your head on the difference between isset and empty.

if(empty($yourVal)) {
  echo "YES empty - $yourVal"; // no result
}
if(!empty($yourVal)) {
  echo "<P>NOT !empty- $yourVal"; // result
}
if(isset($yourVal)) {
  echo "<P>YES isset - $yourVal";  // found yourVal, but result can still be none - yourVal is set without value
}
if(!isset($yourVal)) {
  echo "<P>NO !isset - $yourVal"; // $yourVal is not set, therefore no result
}
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