Re: Simple while loop skipping first row of array

That did the trick, and I learned something new. thanks :) verdon On 2/15/04 11:38 AM, "Duncan" <php-general@ring-wraith.com> wrote: > Problem is, that the first row is already in the $categories variable - > since you have the first mysql_fetch_array call there. > change it to: > > echo "<select name=\"cats\">"; > $result = mysql_query("select * from cats order by title"); > while ($c = mysql_fetch_array($result)) { > echo "<option value=\"$c[id]\">$c[title]</option>"; > } > echo "</select>"; > > > ...and it should work just fine. > > Verdon Vaillancourt wrote: > >> I'm not sure I phrased my subject well, but... >> >> The following is just to build a select menu from a table query result set. >> It sort of works, but doesn't return the first row in the result set, ie if >> I have the following rows in the table 'cats'; >> >> id title >> ========== >> 1 Pears >> 2 Oranges >> 3 Apples >> 4 Bananas >> >> echo "<select name=\"cats\">"; >> $result = mysql_query("select * from cats order by title"); >> $categories = mysql_fetch_array($result); >> if ($categories) { >> while ($c = mysql_fetch_array($result)) { >> echo "<option value=\"$c[id]\">$c[title]</option>"; >> } >> } >> echo "</select>"; >> >> >> The resulting select menu will only include Bananas, Oranges, Pears. >> >> Am I missing something terribly obvious to one less newbie than me? >> >> TIA, >> Verdon >> >> >> > > >