RE: [PHP] quick form question

Depending on how high your error reporting is set, this may or may not give you an error (usually doesn't) <form...> <input type="text" name="firstname" value="<?= $firstname; ?>"> <input type="text" name="lastname" value="<?= $lastname; ?>"> </form> Therefore, when it's first run, all those fields have no value, and will leave the value="" empty... Upon a re-submit, it will return the values entered in the form. There are some more ways to do this for select boxes, radiobuttons and checkboxes... (This is how I do it, others may have a much cleaner, nicer way) <select name="whatever"> <option value="1"<? if ($whatever == "1") { print " selected"; }?>>one</option> <option value="2"<? if ($whatever == "2") { print " selected"; }?>>two</option> </select> I hope this helps! -Mark > I have a .php with a simple form in it to populate a table in a mysql db. > When the user clicks submit, I check to make sure he/she entered the > required info and if all is well, I add the data to teh DB. If > he/she left > out some info, I echo "You left out some info" and I redisplay the form. > When I do this all the info that the user typed originally is gone. Whats > the easiest way to put back what he/she already typed? Thanks