Related
if ($user_id == NULL || $user_name == NULL || $user_logged == NULL) {
$user_id = '-1';
$user_name = NULL;
$user_logged = NULL;
}
if ($user_admin == NULL) {
$user_admin = NULL;
}
Is there any shortest way to do it ?
And if i right, it should be tested with is_null?
It's possible $user_id, $user_name and $user_logged write in one line (maybe array?) without repeating NULL ?
If you want to test whether a variable is really NULL, use the identity operator:
$user_id === NULL // FALSE == NULL is true, FALSE === NULL is false
is_null($user_id)
If you want to check whether a variable is not set:
!isset($user_id)
Or if the variable is not empty, an empty string, zero, ..:
empty($user_id)
If you want to test whether a variable is not an empty string, ! will also be sufficient:
!$user_id
You can check if it's not set (or empty) in a number of ways.
if (!$var){ }
Or:
if ($var === null){ } // This checks if the variable, by type, IS null.
Or:
if (empty($var)){ }
You can check if it's declared with:
if (!isset($var)){ }
Take note that PHP interprets 0 (integer) and "" (empty string) and false as "empty" - and dispite being different types, these specific values are by PHP considered the same. It doesn't matter if $var is never set/declared or if it's declared as $var = 0 or $var = "". So often you compare by using the === operator which compares with respect to data type. If $var is 0 (integer), $var == "" or $var == false will validate, but $var === "" or $var === false will not.
here i have explained how the empty function and isset works please use the one that is appropriate also you can use is_null function also
<?php
$val = 0;
//evaluates to true because $var is empty
if (empty($val)) {
echo '$val is either 0, empty, or not set at all';
}
//evaluates to true because $VAR IS SET
if (isset($val)) {
echo '$val is set even though it is empty';
}
?>
empty() is a little shorter, as an alternative to checking !$user_id as suggested elsewhere:
if (empty($user_id) || empty($user_name) || empty($user_logged)) {
}
To check for null values you can use is_null() as is demonstrated below.
if (is_null($value)) {
$value = "MY TEXT"; //define to suit
}
Please define what you mean by "empty".
The test I normally use is isset().
you can use isset() routine .
also additionaly you can refer an range of is_type () functions like
is_string(), is_float(),is_int() etc to further specificaly test
1.
if(!($user_id || $user_name || $user_logged)){
//do your stuff
}
2 . No. I actually did not understand why you write such a construct.
3 . Put all values into array, for example $ar["user_id"], etc.
<?php
$nothing = NULL;
$something = '';
$array = array(1,2,3);
// Create a function that checks if a variable is set or empty, and display "$variable_name is SET|EMPTY"
function check($var) {
if (isset($var)) {
echo 'Variable is SET'. PHP_EOL;
} elseif (empty($var)) {
echo 'Variable is empty' . PHP_EOL;
}
}
check($nothing);
check($something);
check($array);
Its worth noting - and I only found this out after nearly 9 years of PHP coding that the best way of checking any variable exists is using the empty() function. This is because it doesn't generate errors and even though you turn them off - PHP still generates them! empty() however won't return errors if the variable doesn't exist. So I believe the correct answer is not to check if its null but to do the following
if (!empty($var) && is_null($var))
Note the PHP manual
variable is considered empty if it does not exist or if its value equals FALSE
As opposed to being null which is handy here!
Felt compelled to answer this because of the other responses. Use empty() if you can because it covers more bases. Future you will thank me.
For example you will have to do things like isset() && strlen() where instead you could use empty(). Think of it like this empty is like !isset($var) || $var==false
The best and easiest way to check if a variable is empty in PHP is just to use the empty() function.
if empty($variable)
then
....
I have a question : what is the difference between : $value !='' and !isset($value) ?
The main difference in practical use between $value != '' and !isset($value) is that you can use isset() like this:
if(!isset($value)){
echo '$value is not declared or is equal to null';
}
If $value is not actually set then you won't get a notice with isset(). If you did something like this:
if($value != ''){
echo $value." is not equal to nothing";
}
If $value is not set then this will cause a notice in PHP explaining that $value has not yet been declared.
The thing to note is that isset() will check that the variable has been declared but will not check that it isn't equal to ''.
But what about empty()?
The other part of the puzzle here would be empty() which will check whether a variable is declared and is not equal to ''. You could do something like this:
if(!empty($value)){
echo 'We know that $value is declared and not equal to nothing.';
}
which would be the same as doing this:
if(isset($value) && $value != ''){
echo 'We know that $value is declared and not equal to nothing.';
}
Further reading:
isset() - PHP docs
empty() - PHP docs
isset, empty checks - similar question
Why check both isset and empty?
All of these examples work in PHP 5.3+
$variable != '' simply checks variable against empty string value ( or null because '' == null), if variable is not defined notice will be shown.
!isset($variable) checks whether variable is defined and if it is not returns true, no error/notice will be shown.
But be aware that isset may return false even when variable is defined
$variable = null;
var_dump(isset($variable)); // prints false
To check if variable is truly defined within the current scope, use 'array_key_exists' with 'get_defined_vars'
$variable = null;
var_dump(array_key_exists('value', get_defined_vars())); // prints true
isset() will tell you about the existence of the variable and is not NULL.
empty() can tell you about both, existence of the variable & the value of the variable.
Ex.
$var1 = "";
echo isset($var1);
Output:- true
So to check if the variable was set then use isset() and if you want to check if the value of variable was not a null/empty use empty().
$value != '' means you are checking if $value is not an empty string.
!isset($value) determine if a variable $value is not set and is NULL
Read these links for more information on isset & empty
What's the difference between 'isset()' and '!empty()' in PHP?
http://php.net/manual/en/function.isset.php
http://php.net/manual/en/types.comparisons.php
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.
So I am fairly new to php, but before I really had a solid understanding I picked up this habit. When checking for whether or not a $_SESSION / $_POST / $_GET variable was set I use this:
if($_SESSION['username'] != '' {
//allow access
header('Location: welcomefriend.php')
}
else {
//get out
header('Location: getoutofhereyournotwelcome.php')
}
I have used this in login scripts for checking if the session is set to allow access, etc. So now I know about isset / empty but I always seem to run into problems with those.
So my question is will I ever encounter a problem when using if($some_far != '') to check if a variable is set?
That would show notice if the variable doesnot exist, so do:
if( !empty($_SESSION["username"]) ) {
...
Yes - it'll throw a notice if the index doesn't exist in the array.
Yes, if a form is submitted, but not filled out, $_POST["value"]=="" will give you a different result from isset($_POST["value"]). Also, $_POST["value"]=="" will give you an error on some servers when it is not set.
Checking if $_SESSION['username'] is empty without checking whether there actually is a username entry in the $_SESSION array will throw notices. It's good practice to code around them. One way to do this is something like if (isset($_SESSION['username']) && $_SESSION['username'] != '')
As for empty() vs != "", that's personal preference. One problem you might run into is that empty() does not accept values, only references, so you cannot do empty(somefunction($_SESSION['username'])).
Use empty() if( !empty( $some_far ) ) { to check if a $_POST variable is set and not empty.
If the variable can also be empty but must be set use if( isset( $some_far ) ) {
With isset() the variable also must not be NULL to return true.
You also won't get a notice if the variable is not set as Martin already mentioned in his post.
I always do the following:
if (isset($_SESSION['username']) && $_SESSION['username'] != '') {
//allow access
header('Location: welcomefriend.php')
}
else {
//get out
header('Location: getoutofhereyournotwelcome.php')
}
Testing if the index 'username' exists in the $_SESSION variable first in the if statement will mean that if the index doesn't exist then we won't continue with $_SESSION['username'] != '' thus never causing an error.
Considering
An associative array containing session variables available to the current script
Why not use the predefined function available? array_key_exists
if(array_key_exists('username', $_SESSION) ) {
header('Location: welcomefriend.php');
} else {
header('Location: getoutofhereyournotwelcome.php');
}
I read somewhere that the isset() function treats an empty string as TRUE, therefore isset() is not an effective way to validate text inputs and text boxes from a HTML form.
So you can use empty() to check that a user typed something.
Is it true that the isset() function treats an empty string as TRUE?
Then in which situations should I use isset()? Should I always use !empty() to check if there is something?
For example instead of
if(isset($_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
}