Re: setting Check box
| From: | David Robley | Date: | Wed, 28 Mar 2001 01:58:07 +0000 |
| Subject: | Re: setting Check box | ||
| References: | 1 2 3 | Groups: | php.general |
| Request: | Send a blank email to php-general+get-45809@lists.php.net to get a copy of this message | ||
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;
}
?>