Thanks, that did it.
Steve
----- Original Message -----
From: "David Robley" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Tuesday, March 27, 2001 7:58 PM
Subject: Re: [PHP] setting Check box
> On Wed, 28 Mar 2001 10:18, S. Efurd wrote:
> > This worked for me as well, and helped me with a problem I was having!
> >
> > How would you show what option was selected in a drop down box. I have
> > a form the user fills out, this goes to a confirmation page to ensure
> > the data enetered is correct. I have a drop down box for states, and I
> > cant seem to make it show the state selected it always defaults to
> > first choice?
> >
> > Thanks
>
> I've built a function for displaying drop down lists: yours for free. Oh
> look, Ive even bothered to comment it ;-) It will show selected item[s]
>
> Somebody will no doubt post improvements.
>
> <?php
> /* formDropDown - create an HTML <SELECT>
> * vars: $name - the form variable NAME
> * $value - the SELECTED option which may be an array for multiple
> selected items
> * $labels - assoc. array, list of values=>labels
> * $multiple - if non-empty, select multiple
> * $size - number of rows to show if multiple
> * returns: string, HTML (i.e. for use in echo or print statement) */
>
> function formDropDown($name,$value,$labels,$multiple = '',$size ='') {
> $html = '<SELECT NAME="' .$name . '"';
> if(!empty($multiple)) {
> $html .= ' MULTIPLE';
> }
> if(!empty($size)) {
> $html .= ' SIZE=' . $size;
> }
>
> $html .= ">\n";
> $key = key($labels);
> while($key != "") {
> /* Check if the current value is in the $value variable */
> if(is_array($value)) {
> if(in_array($key,$value)) {
> $selected = ' SELECTED';
> } else {
> $selected = '';
> }
> } else {
> if ($key == $value) {
> $selected = ' SELECTED';
> } else {
> $selected = '';
> }
> }
> $html .= "<OPTION VALUE=\"$key\"$selected>$labels[$key]\n";
> next($labels);
> $key = key($labels);
> }
> $html .= "</SELECT>\n";
> return $html;
> }
> ?>
>
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