How can i compare 2 dates in "Y-m-d" format ? i tried like this :

PHP Code:

$y = date("Y");
$m = date("m");
$d = date("d");
$i = 0;
$ads = mysql_query("select cid,cnd from customers");
while($row = mysql_fetch_array($ads))
{
    $date = $row['cnd'];
    $cid = $row['cid'];
    $y1 = exec("echo $date|cut -d '-' -f 1");
    $m1 = exec("echo $date|cut -d '-' -f 2");
    $d1 = exec("echo $date|cut -d '-' -f 3");
    if(($d1 == $d) || ($d1 > $d))
    {
        if(($m1 == $m) || ($m1 > $m))
        {
            if(($y1 == $y) || ($y1 > $y))
            {
                if($k != '1')
                {
                    show_info($cid);
                    echo "" // html tag;
                }
                else ++$i;
            }
        }
    }
} 


The function returns the number of my clients instead the number of my clients wich have to pay..

if you have to compare two dates ... why don't you transform them into timestamps?
look into strtotime() .... that way it's much easier to compare them.

You can do this in PHP or in SQL before it's parsed back to PHP. If you mean comparing two dates by finding the difference between them (that's what i think you mean) then use this function that returns the days difference between two dates

PHP Code:

function count_days($start, $end) {
    if( $start != '0000-00-00' and $end != '0000-00-00' ) {
        $timestamp_start = strtotime($start);
        $timestamp_end = strtotime($end);

        if( $timestamp_start >= $timestamp_end ) return 0;
        
        $start_year = date("Y",$timestamp_start);
        $end_year = date("Y", $timestamp_end);
        $num_days_start = date("z",strtotime($start));
        $num_days_end = date("z", strtotime($end));
        $num_days = 0;
        $i = 0;
        
            if( $end_year > $start_year ) {
                while( $i < ( $end_year - $start_year ) ) {
                    $num_days = $num_days + gmdate("z", strtotime(($start_year + $i)."-12-31"));
                    $i++;
                }
            }

        return ( $num_days_end + $num_days ) - $num_days_start;
    }
    
    else {
        return 0;
    }
} 


That should work (works fine for me). If you want to know the SQL version, shout up.

Works fine, thank you verry much !