I am not understanding why this is happening with empty php variables - php

I try to do this:
if ($var !== ""){
$message = "whatever";
}
But end up having to do this:
if ($var == ""){
//do nothing
} else {
$message = "whatever";
}
Why does that happen? Shouldn't both of those mean the same thing?

!= and == are opposites (non-strict comparison operators).
!== and === are opposites (strict comparison operators, where the value must match what you are comparing exactly).
If you use != instead of !==, your code should work. But:
You should understand what the actual value of your variable is - it's not an empty string. You can use print_r( $var ); to see it.
It's better to use the strict comparison operators === and !==, because they have well-defined behavior that is easier to remember and debug.

As $var is really string just use:
if ($var){//any non-empty string will work fine as it will be casted to boolean automatically
$message = "whatever";
}

$var could be != '' but not= '' eg .. null
if ($var == ""){
//do nothing
} else {
if (is_null($var) {
$message ='NULL';
} else {
$message = "whatever";
}
}

Related

PHP - Check GET Variables Exist And Not Empty

I want to check the GET variables are not empty, I tried ways but they didn't work.
So I had the code like this:
$u = isset($_GET["u"]);
$p = isset($_GET["p"]);
if ($u !== "" && $p !== "") {
//something
} else {
//do something
}
The I checked the code by sending create.php?u=&p=, but the code didn't work. It kept running the //do something part. The I tried:
echo $u;
echo $p;
It returned 1 and 1. Then I changed it to:
if ($u !== 1 && $p !== 1 && $u !== "" && $p !== "") {
//something
} else {
//do something
}
But it continued to run //do something.
Please help.
You can just use empty which is a PHP function. It will automatically check if it exists and whether there is any data in it:
if(empty($var))
{
// This variable is either not set or has nothing in it.
}
In your case, as you want to check AGAINST it being empty you can use:
if (!empty($u) && !empty($p))
{
// You can continue...
}
Edit: Additionally the comparison !== will check for not equal to AND of the same type. While in this case GET/POST data are strings, so the use is correct (comparing to an empty string), be careful when using this. The normal PHP comparison for not equal to is !=.
Additional Edit: Actually, (amusingly) it is. Had you used a != to do the comparison, it would have worked. As the == and != operators perform a loose comparison, false == "" returns true - hence your if statement code of ($u != "" && $p != "") would have worked the way you expected.
<?php
$var1=false;
$var2="";
$var3=0;
echo ($var1!=$var2)? "Not Equal" : "Equal";
echo ($var1!==$var2)? "Not Equal" : "Equal";
echo ($var1!=$var3)? "Not Equal" : "Equal";
echo ($var1!==$var3)? "Not Equal" : "Equal";
print_r($var1);
print_r($var2);
?>
// Output: Equal
// Output: Not Equal
// Output: Equal
// Output: Not Equal
Final edit: Change your condition in your if statement to:
if ($u != "" && $p != "")
It will work as you expected, it won't be the best way of doing it (nor the shortest) but it will work the way you intended.
Really the Final Edit:
Consider the following:
$u = isset($_GET["u"]); // Assuming GET is set, $u == TRUE
$p = isset($_GET["p"]); // Assuming GET is not set, $p == FALSE
Strict Comparisons:
if ($u !== "")
// (TRUE !== "" - is not met. Strict Comparison used - As expected)
if ($p !== "")
// (FALSE !== "" - is not met. Strict Comparison used - Not as expected)
While the Loose Comparisons:
if ($u != "")
// (TRUE != "" - is not met. Loose Comparison used - As expected)
if ($p != "")
// (FALSE != "" - is met. Loose Comparison used)
You need !empty()
if (!empty($_GET["p"]) && !empty($_GET["u"])) {
//something
} else {
//do something
}
Helpful Link
if ($u !== 1 && $p !== 1 && $u !== "" && $p !== "")
why are you using "!==" and not "!=".
to always simplify your problem solve the logic on paper once using the runtime $u and $p value.
To check if $_GET value is blank or not you can use 2 methods.
since $_GET is an array you can use if(count($_GET)) if you have only u and p to check or check all incoming $_GET parameters.
empty function #Fluffeh referred to.
if($_GET['u']!=""&&$_GET['p']!="")
Hope it helps thx
In you code you should correctly check the variable existence like
if ($u != NULL && $p != NULL && $u != 0 && $p != 0) {
//something
} else {
//do something
}
Wow! I was so dumb... isset returns a boolean. I fixed my problem now. Thank you for answering anyway :)
This fixes:
$u = $_GET["u"];
$p = $_GET["p"];

The best practice for not null if statements

I've been writing my "If this variable is not empty" statements like so:
if ($var != '') {
// Yup
}
But I've asked if this is correct, it hasn't caused a problem for me. Here is the answer I found online:
if (!($error == NULL)) {
/// Yup
}
This actually looks longer than my approach, but is it better? If so, why?
Rather than:
if (!($error == NULL))
Simply do:
if ($error)
One would think that the first is more clear, but it's actually more misleading. Here's why:
$error = null;
if (!($error == NULL)) {
echo 'not null';
}
This works as expected. However, the next five values will have the same and (to many, unexpected) behavior:
$error = 0;
$error = array();
$error = false;
$error = '';
$error = 0.0;
The second conditional if ($error) makes it more clear that type casting is involved.
If the programmer wanted to require that the value actually be NULL, he should have used a strict comparison, i.e., if ($error !== NULL)
It is good to know exactly what is in your variable, especially if you are checking for uninitialized vs null or na vs true or false vs empty or 0.
Therefore, as mentioned by webbiedave, if checking for null, use
$error !== null
$error === null
is_null($error)
if checking for initilized, as shibly said
isset($var)
if checking for true or false, or 0, or empty string
$var === true
$var === 0
$var === ""
I only use empty for ''s and nulls since string functions tend to be inconsistent. If checking for empty
empty($var)
$var // in a boolean context
// This does the same as above, but is less clear because you are
// casting to false, which has the same values has empty, but perhaps
// may not one day. It is also easier to search for bugs where you
// meant to use ===
$var == false
If semantically uninitialized is the same as one of the values above, then initialize the variable at the beginning to that value.
$var = ''
... //some code
if ($var === '') blah blah.
Why just don't
if (!$var)
There are ways:
<?php
error_reporting(E_ALL);
$foo = NULL;
var_dump(is_null($inexistent), is_null($foo));
?>
Another:
<?php
$var = '';
// This will evaluate to TRUE so the text will be printed.
if (isset($var)) {
echo "This var is set so I will print.";
}
?>
To check if it's 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';
}
?>

php "if" condition mystery

I am running into a funny problem with a mischievous "if" condition :
$condition1="53==56";
$condition2="53==57";
$condition3="53==58";
$condition=$condition1."||".$condition2."||".$condition3;
if($condition)
{
echo "blah";
}
else
{
echo "foo";
}
Why does the if condition pass?
Why does php echo "blah"? What do I do to make php evaluate the "if" statement and print "foo"?
The problem here is that you're putting your expressions in strings!
Your $condition1, $condition2, and $condition3 variables contain strings, and not the result of an expression, and the same goes for your $condition variable which will be a string that looks like 53==56||53==57||53==58. When PHP evaluates a string it considers it true if it is not empty and not equal to 0, so your script will output blah.
To fix this you just need to take your expressions out of the strings. It should look like this:
$condition1 = 53 == 56; // false
$condition2 = 53 == 57; // false
$condition3 = 53 == 58; // false
$condition = $condition1 || $condition2 || $condition3; // false || false || false = false
if ($condition) {
echo 'blah';
} else {
echo 'foo'; // This will be output
}
You're evaluating strings as booleans; they'll aways be true (except the strings "" and "0". Get rid of almost all of the quotes in your program.
Those aren't conditions, they're strings.
$condition1=53==56;
$condition2=53==57;
$condition3=53==58;
$condition=$condition1 || $condition2 || $condition3;
if($condition)
{
echo "blah";
}
else
{
echo "foo";
}
Because you're not checking those variables, it's saying if (String) will always return true. (unless "")
You should be doing:
if(53==56 || 53==57 || 53==58)
{
echo "blah";
}
else
{
echo "foo";
}
All $condition* variables will evaluate to true. This is how PHP sees it:
if("53==56" || "53==57" || "53==58")
What you want is this:
$condition1 = 53==56;
$condition2 = 53==57;
$condition3 = 53==58;
It's because you're evaluating a string, and strings other than empty strings evaluate to true.
You are concatting a string together, a non-empty string equals TRUE in php.
Because when the if passes, $condition is a string (a concatenation of) containing the text of your conditions. Try using if(eval($condition)).
String always evaluate to true if its not empty
And btw php make implicit conversion to boolean

Checking if isset and is true?

Is there a function to check both
if (isset($var) && $var) ?
The empty() function will do the job.
Use it with the not operator (!) to test "if not empty", i.e.
if(!empty($var)){
}
You may use the ?? operator as such:
if($var ?? false){
...
}
What this does is checks if $var is set and keep it's value. If not, the expression evaluates as the second parameter, in this case false but could be use in other ways like:
// $a is not set
$b = 16;
echo $a ?? 2; // outputs 2
echo $a ?? $b ?? 7; // outputs 16
More info here:
https://lornajane.net/posts/2015/new-in-php-7-null-coalesce-operator
there you go. that should do it.
if (isset($var) && $var)
if (! empty($var))
It seems as though #phihag and #steveo225 are correct.
Determine whether a variable is considered to be empty. A variable is
considered empty if it does not exist or if its value equals FALSE.
empty() does not generate a warning if the variable does not exist.
No warning is generated if the variable does not exist. That means
empty() is essentially the concise equivalent to !isset($var) || $var
== false.
So, it seems !empty($var) would be the equivalent to isset() && $var == true.
http://us2.php.net/empty
Try the empty function:
http://us2.php.net/empty
isset($a{0})
isset AND len is not 0 seems more reliable to me, if you run the following:
<?php
$a=$_REQUEST['a'];
if (isset($a{0})) { // Returns "It's 0!!" when test.php?a=0
//if (!empty($a)) { // Returns "It's empty!!" when test.php?a=0
echo 'It\'s '.$a;
} else { echo 'It\'s empty'; }
?>
$a = new stdClass;
$a->var_false = false;
$a->var_true = true;
if ($a->notSetVar ?? false) {
echo 'not_set';
}
if ($a->var_true ?? false) {
echo 'var_true';
}
if ($a->var_false ?? false) {
echo 'var_false';
}
This way:
if (($var ?? false) == true) {
}
I am amazed at all these answers. The correct answer is simply 'no, there is no single function for this'.
empty() tests for unset or false. So when you use !empty(), you test for NOT UNSET (set) and NOT FALSE. However, 'not false' is not the same as true. For example, the string 'carrots' is not false:
$var = 'carrots'; if (!empty($var)){print 1;} //prints 1
in fact your current solution also has this type problem
$var = 'carrots'; if (isset($var) && $var){print 1;} //prints 1
as does even this
$var = '1.03'; if (isset($var) && $var == true){print 1;} //prints 1
in fact... if you want to do as you described exactly, you need:
$var = 'carrots'; if (isset($var) && $var === true){print 1;} //Note the 3 Equals //doesn't print 1
I suppose the shortest valid way to test this case is :
if (#$var === true){ print 1;}
But suppressing errors for something like this is pretty awful practice.
Don't know if an exact one already exists, but you could easily write a custom function to handle this.
function isset_and_true($var) {
return (isset($var) && $var == true) ? true : false;
}
if (isset_and_true($a)) {
print "It's set!";
}
Check if the variable is set, and true. Ignore warning message
if(#!empty($foo))

PHP: if(!$one == $two) doesn't work always?

Yes, this is just a question i would like to get an answer on. I experienced it a couple of times, where this:
if(!$one == $two){ echo "Not the same"; }else{ echo "The same"; }
Will not work, and
if($one == $two){ echo "The same"; }else{ echo "Not the same"; }
will work.
Why doesn't it work sometimes? I always need to recode like the second, when the first doesn't work.
! is having higher precedence than == so you should use parenthesis as:
if(!($one == $two))
You need to write
if(!($one == $two))
or
if($one != $two)
since the ! operator has a higher precedence than the == operator.
See also: http://www.php.net/manual/en/language.operators.precedence.php
You need
if(!($one == $two))
This is because without the brackets, it is checking if $one is false and then checking if $two == $one. The following is the only time that it will work without the brackets. Evaluating to if (true == true) as !$one = true.
$one = false;
$two = true;
if (!$one == $two)
{
echo "different";
}

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