//if $myVar is not 0, then print $myVar.
The code above print
1. It works fine.
//if $myVar is not 2, then print $myVar.
I am expecting that the code above print
But it doesn’t print 1.
What’s wrong in the
How can I correctly write for the meaning of “if $myVar is not 2, then print $myVar.”?
if (!$myVar==2) is not “if $myVar does not equal 2”, it is “if !$myVar equals 2”, that is to say “if the boolean negation of $myVar equals 2”.
Since $myVar is 1, which has a boolean value of true, the boolean negation of $myVar is false. Since false doesn’t equal 2 the condition in the if in your second snippet doesn’t validate to true.
What you seem to want is
if ($myVar != 2)
You could throw some parentheses around it though.
$var = 1;
var_dump(! ($var == 2) ); #true
Thank you, ScallioXTX.
Thank you, AnthonySterling.
I think the problem is your “2” is too big
This will also be useful.
In mathematics and mathematical logic, Boolean algebra is a branch of algebra. It differs from elementary algebra in two ways. First, the values of the variables are the truth values true and false, usually denoted 1 and 0, whereas in elementary algebra the values of the variables are numbers. Second, Boolean algebra uses logical operators such as conjunction (and) denoted as ∧, disjunction (or) denoted as ∨, and the negation (not) denoted as ¬. Elementary algebra, on the other hand, uses arith...