Why not check if $thekey is in the $db, then else echo not found?
seems all to much to do so little.
On Friday 06 October 2006 18:35, 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;
> }
> }
> }
>
> 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: [EMAIL PROTECTED]
> http://www.informed-sources.com
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