Are those 2 expressions equivalent (I mean "can I replace the 1st one with the 2nd one):
if ($var) { ... }
and
if (!empty($var)) { ... }
I feel there is a difference but I rationally cannot say which one.
For me the first one evaluates if $var is true or false and I may be wrong but "false" evaluations means that $var is false (boolean), empty (string, object or array), 0 value (int, float or string) or undefined ... that's the way the "empty" function works (http://php.net/manual/en/function.empty.php).
If those tests are equivalent (at least in specific cases), which is better to use (readability, performance, maintenance, ...)?
Thanks
They differ in that for your second example, $var doesn't have to be set before using it. In the first case, if $var isn't set, a notice will be generated, while in the second example, it won't.
This can be useful for values submitted by users inside the $_GET and $_POST superglobals (.. and for $_COOKIE and $_SERVER).
// will generate a notice if there is no `foo` in the query string
if ($_GET['foo'])
// will not generate a notice, even if the key is not set
if (!empty($_GET['foo']))
!empty($var)
Determine whether a variable is considered to be not empty. A variable is considered not empty if it does exist or if its value equals TRUE. empty() does not generate a warning if the variable does not exist.
if ($var) { ... }
You'll test if $var contains a value that's not false -- 1 is true, 123 is too
Extra:
isset($var)
Using isset(), you'll test if a variable has been set -- i.e. if any not-null value has been written to it.
-
It all depends on what you want to check/test. I do hope it helps.
empty() -> If variable not exist or its equals to false empty function returns true.
Imagine that you did not declare $var
if ($var) {
echo '1';
}
else {
echo '2';
}
Output will be:
NOTICE Undefined variable: var on line number *
If you use empty:
if (!empty($var)) {
echo 1;
}
else {
echo 2;
}
Output will be:
2
Also the following values are considered to be empty
$var = 0;
$var = "";
$var = false;
$var = null;
Also check isset() function Php.net isset
In PHP, how do I test whether an environment variable is set? I would like behavior like this:
// Assuming MYVAR isn't defined yet.
isset(MYVAR); // returns false
putenv("MYVAR=foobar");
isset(MYVAR); // returns true
getenv() returns false if the environment variable is not set. The following code will work:
// Assuming MYVAR isn't defined yet.
getenv("MYVAR") !== false; // returns false
putenv("MYVAR=foobar");
getenv("MYVAR") !== false; // returns true
Be sure to use the strict comparison operator (!==) because getenv() normally returns a string that could be cast as a boolean.
you can check like this
if($ip = getenv('REMOTE_ADDR'))
echo $ip;
getenv() Returns the value of the environment variable.
This is what you need
$var = getenv(MYVAR)
if(isset($var)) {
} else {
}
Possibly a strange one that I hope can be done in one line.
I have to have an IF statement that will checks two things.
The first checks if the variable $loggedInfo['status'] is set and is equal to "client".
The second checks that the variable $loggedInfo['address1'] is set and is blank.
The reason being that when the first variable equals staff then the 'address1' variable doesn't exist.
I did have the following but when I log in as staff it still checks for the address1
if((isset($loggedInfo['status'])=="client")&&(!$loggedInfo['address1'])){
//Do something
}
isset returs true or false. you have to do separate check for the actual value
if(
isset($loggedInfo['status']) && $loggedInfo['status']=="client" &&
isset($loggedInfo['address1']) && trim($loggedInfo['address1']) != ''
)
{
//Do something
}
if((isset($loggedInfo['status']) && $loggedInfo['status']=="client") &&(empty($loggedInfo['address1'])){
//Do something
}
isset() returns TRUE if the given variable is defined in the current scope with a non-null value.
empty() returns TRUE if the given variable is not defined in the current scope, or if it is defined with a value that is considered "empty". These values are:
NULL // NULL value
0 // Integer/float zero
'' // Empty string
'0' // String '0'
FALSE // Boolean FALSE
array() // empty array
Depending PHP version, an object with no properties may also be considered empty.
Well you just can't compare the return value of isset() with the string "client", because it will never equal that. To quote http://php.net/manual/en/function.isset.php its return values are "TRUE if var exists and has value other than NULL, FALSE otherwise".
First check if it is set
if ((isset($loggedInfo['status']) === true) && ($loggedInfo['status'] === "client") && (empty($loggedInfo['address1']) === true)) {
// Do something
}
Key take away from this should be to look up return values for every function you use, like empty(), in the manual http://www.php.net/manual/en/function.empty.php. This will save you a lot of headaches in the future.
Considering that:
The isset() construct returns TRUE if a variable is set and not NULL
The is_null() function throws a warning if the variable is not set
Is there a way to test whether a variable exists, no matter it's NULL or not, without using the # operator to suppress the notice?
EDIT
Together with your first replies, I've been thinking about this and I'm getting the conclusion that inspecting get_defined_vars() is the only way to distinguish between a variable set to NULL and an unset variable. PHP seems to make little distinctions:
<?php
$exists_and_is_null = NULL;
// All these are TRUE
#var_dump(is_null($exists_and_is_null));
#var_dump(is_null($does_not_exist));
#var_dump($exists_and_is_null===NULL);
#var_dump($does_not_exist===NULL);
#var_dump(gettype($exists_and_is_null)=='NULL');
#var_dump(gettype($does_not_exist)=='NULL');
?>
$result = array_key_exists('varname', get_defined_vars());
As you already found out, you cannot :
rely on isset, as it return false for a variable that's null.
use $not_exists===null, as it'll raise a notice.
But you could be able to use a combinaison of :
get_defined_vars to get the list of existing variables, including those which are null,
and array_key_exists to find out if an entry exists in that list.
For instance :
$exists_and_null = null;
$exists_and_not_null = 10;
$defined_vars = get_defined_vars();
// true
var_dump(array_key_exists('exists_and_null', $defined_vars)
&& $defined_vars['exists_and_null']===null);
// false
var_dump(array_key_exists('exists_and_not_null', $defined_vars)
&& $defined_vars['exists_and_not_null']===null);
// false
var_dump(array_key_exists('not_exists', $defined_vars)
&& $defined_vars['not_exists']===null);
A couple of notes :
In the first case, the variable exists => there is an entry in the list returned by get_defined_vars, so the second part of the condition is evaluated
and both parts of the condition are true
In the second case, the variable exists too, but is null
which means the first part of the condition is true, but the second one is false,
so the whole expression is false.
In the third case, the variable doesn't exist,
which means the first part of the condition is false,
and the second part of the condition is not evaluated -- which means it doesn't raise a notice.
But note this is probably not that a good idea, if you care about performances : isset is a language construct, and is fast -- while calling get_defined_vars is probably much slower ^^
I would argue here that any code requiring such a comparison would have gotten its semantics wrong; NULL is an unset value in a language that has no straightforward way of distinguishing between the two.
I used a self created function to check this easily, keep in mind it will fire off a PHP warning (I only monitor E_ERROR when I develop).
function isNullOrEmpty( $arg )
{
if ( !is_array( $arg ) )
{
$arg = array( $arg );
}
foreach ( $arg as $key => $value )
{
if( $value == null || trim($value) == "" )
{
return true;
}
}
return false;
}
if (isset($var) && (is_null($var)) {
print "\$var is null";
}
This should do the trick.
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.
Is it true that the isset() function treats an empty string as TRUE?
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']))...
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 / $_POST, for instance.
Now, to work on its value, when you know there is such a value : that is the job of empty.
Neither is a good way to check for valid input.
isset() is not sufficient because – as has been noted already – it considers an empty string to be a valid value.
! empty() is not sufficient either because it rejects '0', which could be a valid value.
Using isset() combined with an equality check against an empty string is the bare minimum that you need to verify that an incoming parameter has a value without creating false negatives:
if( isset($_GET['gender']) and ($_GET['gender'] != '') )
{
...
}
But by "bare minimum", I mean exactly that. All the above code does is determine whether there is some value for $_GET['gender']. It does not determine whether the value for $_GET['gender'] is valid (e.g., one of ("Male", "Female","FileNotFound")).
For that, see Josh Davis's answer.
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.)
If you have a $_POST['param'] and assume it's string type then
isset($_POST['param']) && $_POST['param'] != '' && $_POST['param'] != '0'
is identical to
!empty($_POST['param'])
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.
When and how to use:
isset()
True for 0, 1, empty string, a string containing a value, true, false
False for null
e.g
$status = 0
if (isset($status)) // True
$status = null
if (isset($status)) // False
Empty
False for 1, a string containing a value, true
True for null, empty string, 0, false
e.g
$status = 0
if(empty($status)) // true
$status = 1
if(empty($status)) // False
isset() vs empty() vs is_null()
isset is used to determine if an instance of something exists that is, if a variable has been instantiated... it is not concerned with the value of the parameter...
Pascal MARTIN... +1
...
empty() does not generate a warning if the variable does not exist... therefore, isset() is preferred when testing for the existence of a variable when you intend to modify it...
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($variable) !== FALSE.
I use the following to avoid notices, this checks if the var it's declarated on GET or POST and with the # prefix you can safely check if is not empty and avoid the notice if the var is not set:
if( isset($_GET['var']) && #$_GET['var']!='' ){
//Is not empty, do something
}
$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
isset() tests if a variable is set and not null:
http://us.php.net/manual/en/function.isset.php
empty() can return true when the variable is set to certain values:
http://us.php.net/manual/en/function.empty.php
<?php
$the_var = 0;
if (isset($the_var)) {
echo "set";
} else {
echo "not set";
}
echo "\n";
if (empty($the_var)) {
echo "empty";
} else {
echo "not empty";
}
?>
!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 $array5['name'];
}
//output : {nothing}
?>
Please consider behavior may change on different PHP versions
From documentation
isset() Returns TRUE if var exists and has any value other than NULL. FALSE otherwise
https://www.php.net/manual/en/function.isset.php
empty() does not exist or if its value equals FALSE
https://www.php.net/manual/en/function.empty.php
(empty($x) == (!isset($x) || !$x)) // returns true;
(!empty($x) == (isset($x) && $x)) // returns true;
When in doubt, use this one to check your Value and to clear your head on the difference between isset and empty.
if(empty($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
}