Re: if(), else() problem!

Rahul S. Johari wrote: > Ave, > > code: > > $db = dbase_open("osm.dbf", 0); > if ($db) { > $record_numbers = dbase_numrecords($db); > for ($i = 1; $i <= $record_numbers; $i++) { > $row = dbase_get_record_with_names($db, $i); > if ($row['PHONE'] == $thekey) { > echo ³found²; > } > else { > echo ³not found²; > } > } > } $msg = 'cannot open database!'; if ($db = dbase_open("osm.dbf", 0)) { $msg = 'database contains no records!'; if ($record_numbers = dbase_numrecords($db)) { $msg = 'number not found'; for ($i = 1; $i <= $record_numbers; $row = dbase_get_record_with_names($db, $i),$i++) if ($row['PHONE'] == $thekey) { $msg = 'number found'; break; } } } echo $msg; // ^^^--- untested > > The loop reads each row in the database, and checks whether it matches > $thekey or not. If it does, it prints ³found², if it doesn¹t, it prints ³not > found². But this happens for ³each row² in the database. So if there are 100 > records, and the program does find a match, I¹ll get 99 ³not found² printed, > and one ³found² printed. > > I can easily put an ³exit;² after my echo in the else(), but then it stops > the loop, and doesn¹t go any further. > > What do I have to do to get results if the phone matches, print nothing for > rows where it doesn¹t match, but give one single ³not found² if the phone > number does not exist in the database? > > The logic is just failing me at this point. > > Rahul S. Johari > Supervisor, Internet & Administration > Informed Marketing Services Inc. > 500 Federal Street, Suite 201 > Troy NY 12180 > > Tel: (518) 687-6700 x154 > Fax: (518) 687-6799 > Email: rahul@informed-sources.com > http://www.informed-sources.com > >