<?Php
$x="some value"; // assigned a string to the variable $x here
if(isset($x)){ // checking the existence of the variable by isset
echo "The variable exists";
}else{
echo " Variable does not exists ";}
?>
filter_has_var to check variable
The code above will test the existence of the variable $x and since we have already declared it at first line so the PHP IF condition will return true and the statement inside if condition will be evaluated.<?Php
$test_var=null;
if(isset($test_var)){echo " Output : True";}
else{
echo "Output : False ";
}?>
Output is
Output : False
In the above code, Let us change the variable like this
$test_var='';
The output will change to
Output : True
<?Php
$test_var1='';
$test_var2='';
unset($test_var2);
if(isset($test_var1,$test_var2)){echo " Output : True";}
else{
echo "Output : False ";
}
?>
By using unset() function we have removed one variable hence the output of isset function will be FALSE
<?Php
$my_array=array("first"=>'test', "second"=>34, "third"=>'');
var_dump(isset($my_array['first'])); // bool(true)
var_dump(isset($my_array['second'])); // bool(true)
var_dump(isset($my_array['new_one'])); // bool(false)
?>
$input=array('One' =>1,'Two'=>null);
echo var_dump(isset($input['Two'])); // bool(false)
If the element is equal to NULL then output became False, however array_key_exists() returns True.
$input=array('One' =>1,'Two'=>null);
if(array_key_exists('Two',$input)){ // True
echo "Key is available "; // this is the output
}else{
echo "Key is NOT available ";
}
echo "<br>";
////
echo var_dump(isset($input['Two'])); // bool(false)
if(isset($_SESSION['userid']) && !empty($_SESSION['userid'])) {
echo " session is available, Welcome $_SESSION[userid] ";
}else{
echo " No Session , Please Login ";
exit;
}
Chris Kavanagh | 10-02-2012 |
Very Nice Explanation. This really cleared it up for me. Thank You. |
masroor javed | 30-04-2013 |
thnks for help,,,,, |