Re: if(), else() problem!

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Usen this:

    echo ³not found²;
    break;


""Rahul S. Johari"" <rjohari@xxxxxxxxxxxx> escreveu na mensagem 
news:C14BFE19.193CA%rjohari@xxxxxxxxxxxxxxx
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;
    break;
    }
}
}

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@xxxxxxxxxxxxxxxxxxxx
http://www.informed-sources.com

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