Howcome this condition failes with single or double quotes?

[mikie46]

I'v been going nuts for about 2hrs.

I have the following code;

         $stat2 = `grep -c -m1 "^$user" "$myfile"`;

The above returns either a 0 or 1

Now here is the good part. The if condition fails completely if i use

         if ($stat2 == '0') or if ($stat2 == "0") HUH?

Why is that? Is there a reason?

Intead to get this working i had to place the following condition in my code

if ($stat2 == 0)

Can sombody explain why? I have been using single and double quotes without problems all along and this almost drove me to drink.

[laserlight]

Perhaps $stat2 does not really contain the string '0' (which in this context is equivalent to "0"), but rather a string that is converted to 0. Print it out to check, e.g.,

echo '"' . htmlspecialchars($stat2) . '"';

[HalfaBee]

You probably need

$stat2 = grep -c -m1 "^'.$user.'" "'.$myfile.'";

[mikell]

grep -c -m1 "^'.$user.'" "'.$myfile.'" returns a 0 or 1 of type string
with a trailing newline character "\n". eg. ( "0\n" or "\1\n" )

Knowing that, the following should make a little more sense:

Assume $stat2 = "0\n"
1. ($stat2 == "0") is false
2. ($stat2 == '0') is false
3. ($stat2 == 0) is true

They may seem like odd results, but they are kind of odd comparisons too. (comparing a string to an integer)
In 3, the string "0\n" evaluates to an integer value of 0, so you get (0 == 0), which is true.

Here is a different way to do this you might find useful:

//make sure that $stat2 is an integer
$stat2 = (int) grep -c -m1 "^$user" "$myfile";

if($stat2){
//$stat2 is 1
}
else{
//$stats2 is 0
}