which is better?
if (!empty($val)) { // do something }
and
if ($val) { // do something }
when i test it with PHP 5, all cases produce the same results. what about PHP 4, or have any idea which way is better?
You should use the empty() construct when you are not sure if the variable even exists. If the variable is expected to be set, use if ($var) instead.
empty() is the equivalent of !isset($var) || $var == false. It returns true if the variable is:
"" (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 $var; (a variable declared, but without a value in a class)
Read the manual:
empty() is the opposite of (boolean) var, except that no warning is
generated when the variable is not set.
Related
I have snippet of code which behaviour I do not understand. TRUE value is giving different result when passed to empty() in different copies of variable.
var_dump($this->controller->type['Company']['is_active']); // bool(true)
property $controller is object with $type property which is multi array. value of is_active is bool(true)
What should be the result of empty()?
var_dump(!empty($this->controller->type['Company']['is_active'])); //false
ok, let's create a copy
$temp = $this->controller->type['Company']['is_active'];
var_dump(!empty($temp)); //true
hmmm different result?
var_dump($this->controller->type['Company']['is_active'] === $temp);//true
what about casting?
var_dump(!empty((int)$this->controller->type['Company']['is_active'])); //true
Could someone explain this behaviour please?
Please refere php manual, http://php.net/manual/en/function.empty.php
Returns FALSE if var exists and has a non-empty, non-zero value.
Otherwise returns TRUE.
Note that empty works with pass by variable not with pass by value.
The following things are considered to be empty:
"" (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)
$var; (a variable declared, but without a value)
I also get confused how to check if a variable is false/null when returned from a function.
When to use empty() and when to use isset() to check the condition ?
For returns from functions, you use neither isset nor empty, since those only work on variables and are simply there to test for possibly non-existing variables without triggering errors.
For function returns checking for the existence of variables is pointless, so just do:
if (!my_function()) {
// function returned a falsey value
}
To read about this in more detail, see The Definitive Guide To PHP's isset And empty.
Checking variable ( a few examples )
if(is_null($x) === true) // null
if($x === null) // null
if($x === false)
if(isset($x) === false) // variable undefined or null
if(empty($x) === true) // check if variable is empty (length of 0)
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.
ISSET checks the variable to see if it has been set, in other words, it checks to see if the variable is any value except NULL or not assigned a value. ISSET returns TRUE if the variable exists and has a value other than NULL. That means variables assigned a " ", 0, "0", or FALSE are set, and therefore are TRUE for ISSET.
EMPTY checks to see if a variable is empty. Empty is interpreted as: " " (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), and "$var;" (a variable declared, but without a value in a class.
isset — Determine if a variable is set and is not NULL
$a = "test";
$b = "anothertest";
var_dump(isset($a)); // TRUE
var_dump(isset($a, $b)); // TRUE
unset ($a);
var_dump(isset($a)); // FALSE
empty — Determine whether a variable is empty
<?php
$var = 0;
// 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 '$var is set even though it is empty';
}
?>
It is important to use the correct function / notation, not just whatever appears to work correctly. There are a few things to consider that are not mentioned in the existing answers.
Use isset to check if a variable has either not been set or has been set to null.
Use empty to check if a variable == false. null is cast to false and as with isset, no notice is thrown if the variable has not been set.
if (!$variable) or if ($variable == false) is the same as empty, except that a notice will be thrown if the variable has not been set.
if ($variable !== null) is the same as isset, except that a notice will be thrown if the variable has not been set.
NB
if (!$variable) and if ($variable !== null) perform better than their respective functions but not when notices are being generated, therefore, $variable needs to have been set. Don't suppress notices as a micro-optimisation, as this will make your code harder to debug and even suppressed notices cause a performance penalty.
Coalescing operators
If you are checking a variable so that you can assign a value to it, then you should use ??, ?: instead of if statements.
??
?? assigns a value when not equal to null.
$variable = $a ?? $b is the same as:
if (isset($a))
$variable = $a;
else
$variable = $b;
?:
?: assigns a value when not == to false.
$variable = $a ?: $b is the same as:
if ($a)
$variable = $a;
else
$variable = $b;
but remember that a notice will be generated when $a has not been set. If $a might not have been set, you can use $variable = !empty($a) ? $a : $b; instead.
check false: if ($v === false)
check null: if (is_null($v))
empty() is an evil.It is slow,and when $v queals false,0,'0',array(),'',it will return true.if you need this kind of checking,you can use if ($v).
Are the below two statements similar?
if (isset($someValue) && $someValue != '')
and
if(!empty($someValue]))
No, they are not the same. Yes, they are similar. The first one checks if the 'someValue' is set and not equal to null plus whether it is not empty string. The second one is true if $_GET['act'] is not set or is on the list of variables treated as empty and listed on documentation page.
The following example from the same documentation page should fill any gaps:
$var = 0;
// 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';
}
Kind of similar except 0 value.
According to http://us3.php.net/empty
$var = "0";
(!empty($var)) === false;
(isset($var) && $var!='') === true;
As you write it, they're actually almost the opposite, but I'll just assume you mean isset($_GET[someValue]) and !empty.
In that case no, they're not the same. In the first line you are checking for an empty string, but for example a zero value would pass. However, a zero value counts as empty.
Similar but not the same. Empty will return true if $_GET['act'] is equal to any of the following:
"" (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 $var; (a variable declared, but without a value in a class)
Where as the first one only checks to see if the variable has been created and isn't equal to an empty string.
Code will explain more:
$var = 0;
if (!empty($var)){
echo "Its not empty";
} else {
echo "Its empty";
}
The result returns "Its empty". I thought empty() will check if I already set the variable and have value inside. Why it returns "Its empty"??
http://php.net/empty
The following things are considered to be empty:
"" (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 $var; (a variable declared, but without a value in a class)
Note that this is exactly the same list as for a coercion to Boolean false. empty is simply !isset($var) || !$var. Try isset instead.
I was wondering why nobody suggested the extremely handy Type comparison table. It answers every question about the common functions and compare operators.
A snippet:
Expression | empty($x)
----------------+--------
$x = ""; | true
$x = null | true
var $x; | true
$x is undefined | true
$x = array(); | true
$x = false; | true
$x = true; | false
$x = 1; | false
$x = 42; | false
$x = 0; | true
$x = -1; | false
$x = "1"; | false
$x = "0"; | true
$x = "-1"; | false
$x = "php"; | false
$x = "true"; | false
$x = "false"; | false
Along other cheatsheets, I always keep a hardcopy of this table on my desk in case I'm not sure
In case of numeric values you should use is_numeric function:
$var = 0;
if (is_numeric($var))
{
echo "Its not empty";
}
else
{
echo "Its empty";
}
Use strlen() instead.
I ran onto the same issue using 1/0 as possible values for some variables.
I am using if (strlen($_POST['myValue']) == 0) to test if there is a character or not in my variable.
I was recently caught with my pants down on this one as well. The issue we often deal with is unset variables - say a form element that may or may not have been there, but for many elements, 0 (or the string '0' which would come through the post more accurately, but still would be evaluated as "falsey") is a legitimate value say on a dropdown list.
using empty() first and then strlen() is your best best if you need this as well, as:
if(!empty($var) && strlen($var)){
echo '"$var" is present and has a non-falsey value!';
}
From a linguistic point of view empty has a meaning of without value. Like the others said you'll have to use isset() in order to check if a variable has been defined, which is what you do.
empty() returns true for everything that evaluates to FALSE, it is actually a 'not' (!) in disguise. I think you meant isset()
To accept 0 as a value in variable use isset
Check if variable is empty
$var = 0;
if ($var == '') {
echo "empty";
} else {
echo "not empty";
}
//output is empty
Check if variable is set
$var = 0;
if (isset($var)) {
echo "not empty";
} else {
echo "empty";
}
//output is not empty
It 's working for me!
//
if(isset($_POST['post_var'])&&$_POST['post_var']!==''){
echo $_POST['post_var'];
}
//results:
1 if $_POST['post_var']='1'
0 if $_POST['post_var']='0'
skip if $_POST['post_var']=''
The following things are considered to be empty:
"" (an empty string)
0 (0 as an integer)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but
without a value in a class)
From PHP Manual
In your case $var is 0, so empty($var) will return true, you are negating the result before testing it, so the else block will run giving "Its empty" as output.
From manual:
Returns FALSE if var has a non-empty and non-zero value.
The following things are considered to be empty:
"" (an empty string)
0 (0 as an integer)
"0" (0 as a string) NULL
FALSE array() (an empty array) var
$var; (a variable declared, but
without a value in a class)
More: http://php.net/manual/en/function.empty.php
You need to use isset() to check whether value is set.
Actually isset just check if the variable sets or not.In this case if you want to check if your variable is really zero or empty you can use this example:
$myVar = '';
if (empty($myVar))
{
echo "Its empty";
}
echo "<br/>";
if ($myVar===0)
{
echo "also zero!";
}
just for notice $myVar==0 act like empty function.
if (empty($var) && $pieces[$var] != '0') {
//do something
}
In my case this code worked.
empty should mean empty .. whatever deceze says.
When I do
$var = '';
$var = '0';
I expect that var_dump(empty($var)); will return false.
if you are checking things in an array you always have to do isset($var) first.
use only ($_POST['input_field_name'])!=0 instead of !($_POST['input_field_name'])==0 then 0 is not treated as empty.
Not sure if there are still people looking for an explanation and a solution. The comments above say it all on the differences between TRUE / FALSE / 1 / 0.
I would just like to bring my 2 cents for the way to display the actual value.
BOOLEAN
If you're working with a Boolean datatype, you're looking for a TRUE vs. FALSE result; if you store it in MySQL, it will be stored as 1 resp. 0 (if I'm not mistaking, this is the same in your server's memory).
So to display the the value in PHP, you need to check if it is true (1) or false (0) and display whatever you want: "TRUE" or "FALSE" or possibly "1" or "0".
Attention, everything bigger (or different) than 0 will also be considered as TRUE in PHP. E.g.: 2, "abc", etc. will all return TRUE.
BIT, TINYINT
If you're working with a number datatype, the way it is stored is the same.
To display the value, you need to tell PHP to handle it as a number. The easiest way I found is to multiply it by 1.
proper example. just create int type field( example mobile number) in the database and submit an blank value for the following database through a form or just insert using SQL. what it will be saved in database 0 because it is int type and cannot be saved as blank or null. therefore it is empty but it will be saved as 0. so when you fetch data through PHP and check for the empty values. it is very useful and logically correct.
0.00, 0.000, 0.0000 .... 0.0...0 is also empty and the above example can also be used for storing different type of values in database like float, double, decimal( decimal have different variants like 0.000 and so on.
considering that the variable only can be String fetched from an HTML Form (input-text, textarea).
Yes, there is a difference between strlen($str)==0 and empty($str). empty returns true if the value is "0". See the PHP type comparison tables.
$var = 0;
strlen( $var ); // 1, coerced to true
empty($var) // true, it's considered "empty", these are the empty ones:
"" (an empty string)
0 (0 as an integer)
"0" (0 as a string)
NULL
FALSE
array() (an empty array)
var $var; (a variable declared, but without a value in a class)
Here is some note I've discovered:
empty(), requires a variable and only a variable, so I think it has a problem when it's dealing with an object value fetched from the magic __get() method.