here is what i'm trying to achieve:
if $x is either of these 3 values: 100, 200 or 300 - do something
I'm doing this:
if($x==("100"||"200"||"300"))
{
//do something
}
but //do something is executed even if $x is 400
I noticed that this works:
if($x=="100"||$x=="200"||$x=="300")
{
//do something
}
How is the first block of code different from the second block of code? What am I doing wrong?
The reason why your code isn't working is because the result of the expression:
('100' || '200' || '300')
is always TRUE because the expression contains at least one truthy value.
So, the RHS of the expression is TRUE, while the LHS is a truthy value, therefore the entire expression evaluates to TRUE. The reason why this is happening is because of the == operator, which does loose comparison. If you used ===, the resulting expression would always be FALSE. (unless of course the value of $x is false-y.)
Let's analyze this:
Assuming $x equal '400':
($x == ('100'||'200'||'300'))
// ^ ^
// true true
Make sense now?
Bottom line here is: This is the wrong way of comparing 3 values against a common variable.
My suggestion is that you use in_array:
if(in_array($x, array('100', '200', '300')) {
//do something...
}
you can take all values in array it is working Perfectly.
$x=400;
if(in_array($x, array('100', '200', '300'))) {
echo $x.'is in array';
} else {
echo $x.'is not in array';
}
