hi ,
I have a general question which im not sure of.
when i write this code:
<?php
$abc = true;
if ($ver == $var)
{
$abc = false;
}
if (!$abc)
{
echo "something";
}
else
{
echo "nothing";
}
?>
Now , my problem comes to the statement if (!$abc) :
does this ask : if $abc = false or does it check what $abc currently is (false or true) and requires the opposite?
If my question is anyhow unclear tell me and ill rephrase.
thanks ahead guys .
if (!$abc)
asks if $abc = false then do ....
To better explain it "!" means opposite. So in general if you test for a variable, you say if($var) meaning that if $var equates to true in one of three ways: (1) Is a string (empty or not), (2) is a number not equal to 0, or (3) is the boolean "true". So by saying !$var you're saying If $var != true....
bpat1434 wrote:To better explain it "!" means opposite. So in general if you test for a variable, you say if($var) meaning that if $var equates to true in one of three ways: (1) Is a string (empty or not), (2) is a number not equal to 0, or (3) is the boolean "true". So by saying !$var you're saying If $var != true....
So making sure i got it right - if i set:
$var = false;
if (!$var) would mean:
if $var != false;
which would mean:
if $var = true ;
correct ? 
No it mean
if $var != true;
or
if $var = false
No... $var would still be false, but what would execute is that first if statement. Think of it this way:
!$var is the same as saying $var == false
and
$var is the same as saying $var == true
Alright got it , thanks guys =]
