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  1. #1
    Regular Coder
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    Question higher number check fails

    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?


    Code:
    $num_rows = mysql_num_rows($resultaat)+1;
    
    echo $code, $email,$num_rows,$numbers;
    
    if ($num_rows>$numbers){echo "max reached.";exit;}

  • #2
    UE Antagonizer Fumigator's Avatar
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    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

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

  • #3
    Master Coder
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    hmmm

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

    $num_rows 
    mysql_num_rows($resultaat)+1;

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

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


    .


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