Re: dynamic form help

I would create an array of the options then loop through them and checking if the current option is the same as the current selection. Let's take your example. $array_options[] = 1; $array_options[] = 2; $array_options[] = 3; echo "<SELECT>\n"; for ($j = 0; $j < count($array_options); $j++) { echo "\t<OPTION"; if ($current_val == $array_options[$j]) { echo " selected"; } echo ">" . $array_options[$j] . "</OPTION>\n"; } echo "</SELECT>\n"; This means that adding another option is easy and if $current_val is set then it will always select the correct option. This can be expanded for multiple selects as well. Jay ----- Original Message ----- From: "Jon Rosenberg" <php@columbusinteractive.com> To: "PHP List" <php-general@lists.php.net> Sent: Wednesday, November 22, 2000 3:51 PM Subject: Re: [PHP] dynamic form help > Thanks for that bit of help...but, is there any way to do it where I don't > have the list of options stored in the db? I can add them to the db, but > I'd rather not since I have so many form fields to do. Thanks!!! > > > ----- Original Message ----- > From: "Cristian Donoso" <cdonoso@woc.cl> > To: "Jon Rosenberg" <php@columbusinteractive.com>; "PHP List" > < > > Sent: Wednesday, November 22, 2000 3:27 AM > Subject: RE: [PHP] dynamic form help > > > > > What would be the best way to do the following? I have a mysql db > record, > > I > > > have it so people can login against the db and now I want to allow them > to > > > edit the information about them in the db. The form they create their > > entry > > > with is a combination of text boxes and drop-down selection menus. When > > > they log in I want the form that comes up to have the values in it that > > they > > > originally signed up with. I know how to populate the text box fields > > > dynamicly with value=$var_name , but how can I make it so that the > > selection > > > drop-down menus have what they selected? I essentially need to be able > to > > > write those parts of the form like this: > > > > > > drop-down: > > > > > > You have 3 choices, your current selection is...you can change it if > you'd > > > like and save it to the db: > > > <SELECT> > > > <OPTION>1</OPTION> > > > <OPTION selected>2</OPTION> > > > <OPTION>3</OPTION> > > > </SELECT> > > > > > > how can I make it so that the 'selected' appears in the right place? > and > > > yet still include all the other options so they can change it if they > want > > > to? > > > > If you have stored in your DB the option, you can do something like this: > > > > print "<SELECT>"; > > for ($i=1 ; $i<=3; $i++) { > > ($i == $value_from_database) ? $sel = "selected" : $sel = ""; > > print "<OPTION $sel>$i</OPTION>"; > > } > > print "</SELECT>"; > > > > this only works if the $value_from_database is an INTEGER but that's the > > idea. > > > > ______________________________ > > Cristian Donoso Monteiro > > Temuco - Chile > > > > > > > > -- > > PHP General Mailing List (http://www.php.net/) > > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > > For additional commands, e-mail: php-general-help@lists.php.net > > To contact the list administrators, e-mail: php-list-admin@lists.php.net > > > > > -- > PHP General Mailing List (http://www.php.net/) > To unsubscribe, e-mail: php-general-unsubscribe@lists.php.net > For additional commands, e-mail: php-general-help@lists.php.net > To contact the list administrators, e-mail: php-list-admin@lists.php.net >