close
$query = mysql_query("SELECT root FROM training");
echo "<form action=something.php method=POST><select name='Field'>";
if($mysql_num_rows($result)>0){
while ($r = mysql_fetch_array($query))
{
$user = $r["username"];
echo "<option value='$user'>$user</option>";
}
}
echo "</select>";
should load the data in the select.
To link select boxes, first you need to decide on a method:
1. ajax - fire an ochange event to do an ajax call to load the next select
box
2. pass the other select box data down to a javascript array and let the
choices from the first select choose which array gets loaded into the next
select
hth
Bastien
From: Scott <[EMAIL PROTECTED]>
To: php-db@lists.php.net
Subject: [PHP-DB] Creating drop down lists
Date: Sat, 10 Mar 2007 04:33:37 -0700
Hi everyone!
I would like to create a drop down list with the info in my database.
I would also like to have more options so it will filter down to many
rows of data displayed then filtered down to a few rows of data that
said user is looking for.
Kind of like select a state then in the next box you can select the city.
but first thing is first.....
I have tried a few things out there that I searched for but everything
I tried it will not populate the drop down list with the data in my
database.
So I know it somewhere along the lines of:
$query = mysql_query("SELECT root FROM training");
echo "<form action=something.php method=POST><select name=Field>";
while ($r = mysql_fetch_array($query))
{
$user = $r["username"];
echo "<option value=$user>$user</option>";
}
echo "</select>";
or something like that but nothing populates....
What is the proper code to get this done to populate and display what
I select from the drop down menu?
Say my DB name is: TUBA
My table name is: FLUTE
and I have xxx amounts of rows to filter through with three columns in
my mysql db.
Does that all make sense?
Thanks in advance!
You all rock!!
sb
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