PHP & multiple drop-down lists

Ted_Haller

Refering to an earlier post of mine regarding populating drop-down lists according to what was selected in a previous one, it was suggested I use javascript to populate arrays and then use these to populate the drop-down lists.

It is not practical for me to do this, one reason being that the four lists I'm using would be populated from 4 different tables which are updated quite frequently.

So I've decided to reload the page (which is mostly text so it shouldn't take too long) after each of the first three selections. The problem I'm having now is that when I reload the page, php doesn't seem to see what was selected! It is probably something really stupid but I just can't see it.

What I've got is two forms on the page. The first contains the four '<select>...</select>' elements and an image being used as the submit button. The form tab looks like this:

<form style="margin-bottom:0;" name="fruitspec" action="<? echo $PHP_SELF ?>?action='fruitspec'" method=post>

The 1st drop down list is populated (this works as I get all the correct options):

...db connection etc...
while( $row1 = mysql_fetch_array( $res1 ))
{
$code = $row1[0];
$desc = $row1[1];
$option_block .= "<option value = \"$code\">$code - $desc</option>";
}
?>
<select name="commgrp"><? echo $option_block ?></select>

The submit button is as follows:

The rest of the document then consists of another form used to capture the rest of the data.

When the page reloads, '$commgrp' is blank. I'm stumped! Does php not like multiple forms?? I had started out using nested forms which gave similar results.

Any suggestions or tips would be GREATLY appreciated.

Thanks

Ted

merkinmuffley

hi ted. first, you can't submit two different forms at the same time to the same script. one form submission = one request, which is all the technology can handle at one time. i'm not sure if that's what you are trying but it seems so.

next- let's assume for a second that your form only has 1 select widget called commgrp. the form submits to PHP_SELF, so upon reload the select box needs to be defaulted to what the user just selected. there is only 1 thing that needs to happen- put the SELECTED keyword into the correct OPTION tag. the code you posted doesn't do this. all it needs to do is, inside the while loop, check whether ($code == $commgrp) and if so put the keyword inside the option tag. just extend that concept to however many select widgets you need.

and just in case that doesn't work, remember that the configuration setting register_globals needs to be turned on in order for the variable $commgrp to appear when the form is submitted. if that setting is turned off, you have to use $_POST["commgrp"] instead. (which, fwiw, is a more secure route to go.)

/nick

waygood

I think multiple forms is okay but don't nest them, set them all to POST to $_SERVER['PHP_SELF'] except the last one that process the data.

<form name="one">
if isset($_POST['drop_downbox1']) add in SELECTED in correct postion
drop_downbox1
</form>

<form name="two">
<input type="hidden" name="drop_downbox1" value="<php echo $_POST['drop_downbox1']; ?>" />
if isset($_POST['drop_downbox2']) add in SELECTED in correct postion
drop_downbox2
</form>

<form name="three">
<input type="hidden" name="drop_downbox1" value="<php echo $_POST['drop_downbox1']; ?>" />
<input type="hidden" name="drop_downbox2" value="<php echo $_POST['drop_downbox2']; ?>" />
if isset($_POST['drop_downbox3']) add in SELECTED in correct postion
drop_downbox3
</form>


<form name="maindata">
<input type="hidden" name="drop_downbox1" value="<php echo $_POST['drop_downbox1']; ?>" />
<input type="hidden" name="drop_downbox2" value="<php echo $_POST['drop_downbox2']; ?>" />
<input type="hidden" name="drop_downbox3" value="<php echo $_POST['drop_downbox3']; ?>" />
rest of it here
<input type="image" src="../images/buttons/accept_MOff.gif" border=0 width=38 height=10 alt="Accept">
</form>

if you don't want to show them all at once just wrap each one in a test for previous drop_downbox value :-

if (isset($_POST['drop_downbox2'])
{
<form name="three">
<input type="hidden" name="drop_downbox1" value="<php echo $_POST['drop_downbox1']; ?>" />
<input type="hidden" name="drop_downbox2" value="<php echo $_POST['drop_downbox2']; ?>" />
if isset($_POST['drop_downbox3']) add in SELECTED in correct postion
drop_downbox3
</form>
}

a submit will only send data that is within the FORM tags surrounding it, as long as they are not nested.
This is because HTML can't cope with data within multiple forms

<form>
anything form related here is ignored
<form>
this bit ok
</form>
anything here ignored including submit
<form>

html says:-
start form
oops! I meant start form
do form stuff
end form
don't need to end a form cos I already did it