I use this function:
function ListboxMatch($size, $name, $query, $matchtothis)
{
//echo "<BR>".$query."<BR>";
if ( $qry = mysql_query($query) )
{
if (mysql_num_rows($qry) > 0)
{
echo "<SELECT SIZE='".$size."'
NAME='".$name."'>\n\t<OPTION VALUE=''>\n";
while (list($value, $text, $description)
= mysql_fetch_row($qry))
{
echo "\t<OPTION VALUE='".$value."'";
if ($value == $matchtothis) {
echo " SELECTED"; }
echo ">";
echo stripslashes($text);
if ($description) echo "
(".stripslashes($description).")";
echo "\n";
}
echo "</SELECT>";
}
else echo "No data for Listbox\n";
mysql_free_result($qry);
}
else echo "Listbox cannot be built because of an
invalid SQL query.";
} /* end ListboxMatch */
And use it like this:
<?php ListboxMatch(1, "edit_booth_id", "SELECT booth_id, booth_name,
booth_number FROM Booth_Table ORDER BY booth_name, booth_number",
$edit_booth_id); ?>
DÆVID.
http://daevid.com
> -----Original Message-----
> From: Natividad Castro [mailto:[EMAIL PROTECTED]]
> Sent: Thursday, May 30, 2002 1:53 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP-DB] drop down menu
>
>
> Hi to all,
> how can I display information in a drop down menu?
>
> What I want to do is: select all the values from info_sys
> field and display them in a drop down menu.
>
> This is what I have so far, but only display one value at a
> time. $sql = mysql_query("SELECT info_sys from project order
> by project_id, project_id ASC"); while($query_data =
> mysql_fetch_array($sql)) { $info_sys = $query_data[info_sys];
> } <select name="info_sys"> <option><? echo $info_sys;
> ?></option> </select>
>
> Any help is greatly appreciate
>
> Thanks in advanced
> Nato
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