View Full Version : higher number check fails

11-15-2012, 12:31 PM
I'm kinda puzzled by this..

- I retrieve a value of a db field and store it into the variable $numbers.
- I count the number of records in an other database table and store that value in $num_rows.

If the value in $num_rows is higher then $numbers the user gets a warning.

My problem is that the warning is allways displayed even if the value of $num_rows is lower then $numbers.

For example, the coding below displays this:
test mymail@live.nl15max reached.

($num_rows is 1, and $numbers is 5 in this case)

Obviously the higher number check fails, since $num_rows is less then $numbers. How to solve this issue?

$num_rows = mysql_num_rows($resultaat)+1;

echo $code, $email,$num_rows,$numbers;

if ($num_rows>$numbers){echo "max reached.";exit;}

11-15-2012, 03:33 PM
From your echo statement it's possible $num_rows is = 15 and $numbers is null. Since we can't see where $numbers got assigned a value I'd say this may be what's happening. If you wrote your echo statement a bit more clearly then you might see what's really going on-- obviously 1 is not greater than 5 and PHP knows that :p

Try using var_dump() on the two variables-- that will tell you if the variable types are numeric or strings.

11-15-2012, 03:38 PM

See what this does:
just wondering if it's an issue of syntax (or formation of the "if" statement) ...

$num_rows = mysql_num_rows($resultaat)+1;

echo $code, $email,$num_rows,$numbers;

if($num_rows > $numbers){
echo "max reached.";
echo "ok";