$a = 101;
if (isset($a) && is_int($a) && in_array($a, range(1, 100))) {
echo "TRUE";
} else echo "FALSE";
Why this condition returns FALSE as it should be while this IF:
if (isset($argv[1]) && is_int($argv[1]) && in_array($argv[1], range(1, 100))) {
echo "TRUE";
} else echo "FALSE";
returns also FALSE where value passed as first parameter is 50 which is in range??? PHP-CLI is 7.0.9-TS-VC14-x64
Thanks in advance
argv[1] is, by default, an string.
Use is_numeric() instead of is_int(), and convert (or cast) it to integer value before checking if in the range.
// var_dump($argv); // ...if you want to check $argv types and values.
if (isset($argv[1]) && is_numeric($argv[1]) && in_array(intval($argv[1]), range(1, 100))) {
echo "TRUE";
}
else {
echo "FALSE";
}
CAUTION: is_numeric() also returns TRUE in case of a float value!
Related
Here is my sample code:
$issue_id = $_POST['issue_id'];
if(!empty($issue_id)){
echo 'true';
}
else{
echo 'false';
}
If I pass 0 to $_POST['issue_id'] by form submitting then it echo false. Which I want is: Condition will be true if the following conditions are fulfilled:
1. true when I pass any value having 0.
2. false when I don't pass any value. i.e: $_POST['issue_id'] is undefined.
I also tried this:
if(!isset($issue_id)){
echo 'true';
}
else{
echo 'false';
}
if(!empty($issue_id) || $issue==0){
echo 'true';
}
else{
echo 'false';
}
The last one is okay, meaning if I pass any value having ZERO then it will echo true. But it will also echo true if I don't pass any value. Any idea?
The last is okay, meaning if I pass any value having ZERO then it echo true. But it also echo true if I don't pass any value. Any idea?
if (isset($_POST["issue_id"]) && $_POST["issue_id"] !== "") {
}
please notice I used !== not !=. this is why:
0 == "" // true
0 === "" // false
See more at http://php.net/manual/en/language.operators.comparison.php
also if you are expecting number you can use
if (isset($_POST["issue_id"]) && is_numeric($_POST["issue_id"])) {
}
since is_numeric("") returns false
http://php.net/manual/en/function.is-numeric.php
Alternatively if you expect number good option is filter_var
if (isset($_POST["issue_id"]) {
$issue_id = filter_var($_POST["issue_id"], FILTER_VALIDATE_INT);
if ($issue_id !== false) {
}
}
since filter_var("", FILTER_VALIDATE_INT) will returns false and filter_var("0", FILTER_VALIDATE_INT) will return (int) 0
http://php.net/manual/en/function.filter-var.php
if(isset($_POST['issue_id'])) {
if($_POST['issue_id'] == 0) {
echo "true";
}
else {
echo "false";
}
}
When you get data from a form, remember:
All text boxes, whether input or textarea will come as strings. That includes empty text boxes, and text boxes which contain numbers.
All selected buttons will have a value, but buttons which are not selected will not be present at all. This includes radio buttons, check boxes and actual buttons.
This means that $_POST['issue_id'] will be the string '0', which is actually truthy.
If you need it to be an integer, use something like: $issue_id=intval($_POST['issue_id']);
#Abdus Sattar Bhuiyan you can also full fill your two condition like below one:
<?php
$_POST["issue_id"] = "0";
$issue_id = isset($_POST['issue_id']) ? (!empty($_POST['issue_id']) || $_POST['issue_id'] === 0 || $_POST['issue_id'] === "0") ? true : false : false;
if($issue_id){
echo 'true';
}
else{
echo 'false';
}
I have $first and $second. They can have value 0 or 1.
if ( $first AND $second ) {
// True
} else {
// False
}
My mind (and Google search) tells me, that the result is true only when $first == 1 and $second == 0 or vice versa. But the result is true when both of this variables are 1.
I don't understand how does it works.
Your Google searches have failed you. PHP's type juggling means that a 1 is equivalent to TRUE and 0 is equivalent to FALSE. (See also type comparisons). So if both values are 1 then that if statement evaluates to TRUE. If one or both values are 0, it evaluates to FALSE.
<?php
$one = 1;
$zero = 0;
if ($one && $one) {
echo "true\n";
}
else {
echo "false\n";
}
if ($zero && $zero) {
echo "true\n";
}
else {
echo "false\n";
}
if ($one && $zero) {
echo "true\n";
}
else {
echo "false\n";
}
if ($zero && $one) {
echo "true\n";
}
else {
echo "false\n";
}
Program Output
true
false
false
false
Demo
In PHP all values are either "truthy" or "falsy" in expressions.
If a value contains something then it can be said to be truthy. So, values such as 1, "one", [1,2,3] or true all "contain" something and will be interpreted as truthy.
Values that are zeroed or in some way empty are falsy. E.g. 0, "", [] and false.
There is a table of how values are interpreted in the PHP documentation.
You can also just experiment, and output it to your website:
var_dump(1 and 0);
I have a variable that can be int or bool, this is because the db from where im querying it change the variable type at some point from bool to int, where now 1 is true and 0 is false.
Since php is "delicate" with the '===' i like to ask if this is the correct why to know if that var is true:
if($wallet->locked === 1 || $wallet->locked === true)
I think in this way im asking for: is the type is int and one? or is the var type bool and true?
How will you approach this problem?
Your code is the correct way.
It indeed checks if the type is integer and the value is 1, or the type is boolean and the value is true.
The expression ($x === 1 || $x === true) will be false in every other case.
If you know your variable is an integer or boolean already, and you're okay with all integers other than 0 evaluating to true, then you can just use:
if($wallet->locked) {
Which will be true whenever the above expression is, but also for values like -1, 2, 1000 or any other non-zero integer.
$wallet->locked = 1;
if($wallet->locked === true){
echo 'true';
}else{
echo 'false';
}
will produce:
false
and
$wallet->locked = 1;
if($wallet->locked == true){
echo 'true';
}else{
echo 'false';
}
will produce:
true
Let me know if that helps!
Your solution seems to be perfect, but You can also use gettype. After that You can check the return value with "integer" or "boolean". Depending on the result You can process the data the way You need it.
solution #1. If $wallet has the value of either false or 0, then PHP will not bother to check its type (because && operator is short-circuit in PHP):
$wallet = true;
//$wallet = 1;
if( $wallet && (gettype($wallet) == "integer" || gettype($wallet) == "boolean") )
{ echo "This value is either 'true and 1' OR it is '1 and an integer'"; }
else { echo "This value is not true"; }
solution #2 (depending on what You want to achieve):
$wallet = 0;
//$wallet = 1; // $wallet = 25;
//$wallet = true;
//$wallet = false;
if($wallet)
{ echo "This value is true"; }
else { echo "This value is not true"; }
This is the code
$a = 'Rs 15.25';
if ( $a != '' && $a! = 0 ) {
echo "Inside If";
} else {
echo "Outside If";
}
actually I want to Print "Inside If" so that's why I put $a='Some String Value'. But it always prints "Outside If". Then I changed my code to
$a = 'Rs 15.25';
if ( $a != '' && $a != '0' ) {
echo "Inside If";
} else {
echo "Outside If";
}
I have just added single quotes to 0. Then i got the exact output as i want. But I didn't understand why this happens.
So please help me with this.
PHP does weak type comparison, that is, it converts both operands to the same type before doing the actual comparison.
If one of the operands is a number, the other one is converted to a number as well. If the second operand is a string and contains no digits, it is silently converted to the number 0.
To avoid this whole issue, use string type checking with the operator !== (=== for equality).
if($a !== '' && $a !== 0) {
echo "Inside If";
} else {
echo "Outside If";
}
First of all you when you have multiple conditions on an if statement you should always enclose each of them within brackets
So first thing you should do is to change your code to
$a='Rs 15.25';
if(($a!='') && ($a!='0'))
{
echo "Inside If";
}else
{
echo "Outside If";
}
In PHP 0 = FALSE, 1 = TRUE.
if($a != 0) -> if($a != FALSE)
if $a = 'Rs 15.25', $a != false and $a not empty, then you have echo "Outside If";
Is there a function to check both
if (isset($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))