Re: Problem with array

Daniel, This is how I did what you are trying to do: ____start____ <form action="province.php3" method="get"> <select name="provid"> <?php $result99 = mysql_query("SELECT * FROM province ORDER BY name",$db); if ($myrow = mysql_fetch_array($result99)) { do { $name=$myrow["name"]; printf("\t<option value=%s>%s\n",$myrow["provid"],$myrow["name"]); } while ($myrow = mysql_fetch_array($result99)); } else{ echo "no records"; } ?> </select> <br> <input type="image" border="0" src="graphics/buttons/go.gif" alt="Go!" value="Go!"> </form> ____ends____ It seems to work fine HTH Stuart ----- Original Message ----- From: Daniel Rezny <dr@photo.digitallaut.at> To: php-db <php-db@lists.php.net> Sent: Wednesday, August 23, 2000 9:41 PM Subject: [PHP-DB] Problem with array > Hello all, > > I have a very interesting problem with mysql and php. > I want to make a html <select> tag with dynamicaly generated <option>. > It works but not correct. > This is the code > $exec = mysql_query("select id,bezeichnung from kategorien where user='".$userid."'"); > > $count = mysql_num_rows($exec); > echo $count; > > $katid=$row[0]; > $katname=$row[1]; > > > $i = 1; > > > for($i=1; $i<=$count; $i++) > > { > $line = mysql_fetch_array($exec, $i); > echo ?><option value="<? echo $line["id"];?>"><?echo $line["bezeichnung"];?></option><?; > } > > > So and it is not writes every row from db but every second row. Do > somebody know where is the problem. > > Thx for every help. > > -- > Best regards, > Daniel mailto:dr@photo.digitallaut.at > > > > -- > PHP Database Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-db-unsubscribe@lists.php.net > For additional commands, e-mail: php-db-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net > >