Re: logic operands problem

Hi everybody, thank you for all the help and thoughts. I have solved it, but I guess it is not an elegant solution. What I do now, is simply check again for the second case. There are 2 cases. Either first page OR all pages, second case: following pages OR all pages. My booking checking looks now as following: ################################################################################################ # on which page will the tl be placed? if ($data[page] == 3){ // all pages

       $where_page = 'AND (page = 1 OR page = 3)'; // unfortunatelly we have to test later on page = 2 OR page = 3. No solution inside one query. We tried also (page=1 XOR page=2) OR page = 3
       $where_page_2 = 'AND (page = 2 OR page = 3)';

} else{ // page one or all following pages

       $where_page = 'AND page = '.$data[page];

} ################################################################################################ ################################################################################################ # find out first possible booking period do{

       // get all toplistings that are at least with one day inside the desired booking period
       $stmt= "
           SELECT
               *
           FROM
               $DB.$T54
           WHERE
               cat_id = '$data[cat_id]'
               AND cat_type = '$data[cat_type]'
               $where_page
               AND
               (
                   (start_date <= '$new_start' AND expires >= '$new_start')
                   OR (start_date <= '$new_end' AND expires >= '$new_end')
                   OR (start_date >= '$new_start' AND start_date<= '$new_end')
               )                      ";
       #echo $stmt.$br;
       $result = execute_stmt($stmt, $link);
       while ($row = db_get_row($result)){              $booked[start][]     = $row->start_date;
           $booked[end][]        = $row->expires;
       }
       $possible_bookings = count($booked[start]);
       // would there be more bookings then possible?
       if ($possible_bookings >= $places){ // not enough space. Try nest day
           $shift = true; // shift period for one day
           $reservation = 1;
           $new_start     =  date("Ymd",strtotime($new_start." + 1 day"));
           $new_end     = date("Ymd",strtotime($new_end." + 1 day"));
       }
       else{ // enough space
           unset($shift);
       }
       unset($booked);      } while ($shift); // shift as long as we find free space
   ################################################################################################    

################################################################################################ # if client wants to book all pages, we have to try also the second constellation # we could not find a way to do this in one sql query # find out if booking period has to be shifted even further due to all pages booking if ($page == 3){

       do{
           // get all toplistings that are at least with one day inside the desired booking period
           $stmt= "
               SELECT
                   *
               FROM
                   $DB.$T54
               WHERE
                   cat_id = '$data[cat_id]'
                   AND cat_type = '$data[cat_type]'
                   $where_page_2
                   AND
                   (
                       (start_date <= '$new_start' AND expires >= '$new_start')
                       OR (start_date <= '$new_end' AND expires >= '$new_end')
                       OR (start_date >= '$new_start' AND start_date<= '$new_end')
                   )                          ";
           //echo $stmt.$br;
           $result = execute_stmt($stmt, $link);
           while ($row = db_get_row($result)){                  $booked[start][]     = $row->start_date;
               $booked[end][]        = $row->expires;
           }
           $possible_bookings = count($booked[start]);
           // would there be more bookings then possible?
           if ($possible_bookings >= $places){ // not enough space. Try nest day
               $shift = true; // shift period for one day
               $reservation = 1;
               $new_start     =  date("Ymd",strtotime($new_start." + 1 day"));
               $new_end     = date("Ymd",strtotime($new_end." + 1 day"));
           }
           else{ // enough space
               unset($shift);
           }
           unset($booked);          } while ($shift); // shift as long as we find free space

}

################################################################################################      
This is rather a dirty solution. Maybe someone has an idea on how to do it more elegant? I believe there should be one simple line that is different instead of checking it all over again for a second time.

Any ideas? Kim Madsen wrote:

Hey Merlin Merlin Morgenstern wrote on 2009-12-07 11:52:

Hello everybody, I am having trouble finding a logic for following problem: Should be true if: page = 1 OR page = 3, but it should also be true if page = 2 OR page = 3 The result should never contain 1 AND 2 in the same time. This obviously does not work: (page = 1 OR page = 3) OR (page = 2 OR page = 3) This also does not work: (page = 1 OR page = 3 AND page != 2) OR (page = 2 OR page = 3 AND page != 1) Has somebody an idea how to solve this?

I've read the entire thread and can see that this is a MySQL query you want to make (I was about to tell you about the == comparison and the $ in a variable in PHP). What you want is all results containing 1,2 or 3, so make a "WHERE page IN(1,2,3)" and use PHP logic to figure out if a free slot is available or not. Or rewrite your booking routine (use data and time fields instead, maybe by creating a bunch of free slots and then a booked field with a default of "0", changed to "1" when booked)