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