hi,
I am using the following code, but it is not populating my script. If you can help
I would be grateful.
I am using mysql as database.
<?
$sql = "SELECT course_id FROM course";
$sql_result = mysql_query($sql)
or die("Couldn't execute query.");
while ($row = mysql_fetch_array($sql_result)) {
$type = $row["course_id"];
$typedesc =$row["dept_id"];
$option_block .= "<OPTION value=\"$type\">$typedesc</OPTION>";
}
?>
<SELECT name="selecttype" id="selecttype">
<? echo "$option_block"; ?>
</SELECT>
thanks.
"John W. Holmes" <[EMAIL PROTECTED]> wrote:
From: "andy amol"
> I would like to know how to create and populate drop down boxes in php.
> I want the value to be populated from database.
> What I am try to do is to provide the forign key value as combo box
option, so that I do not have to check for referential integrity.
You still have to check. Just because you provide a discreet number of
options in a box doesn't mean that's really all the user can choosefrom. There are
many ways to manipulate the data.That being said, just create a loop as you draw items
from your database.echo '';
$sql = "SELECT name FROM products WHERE ...";
$result = query($sql);
while($row = fetch_assoc($result))
{ echo "{$row['name']}\n"; }
echo "";
I don't know what database you're using, so query() and fetch_assoc() are
generic.
---John Holmes...
--
PHP Database Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
---------------------------------
Do you Yahoo!?
Yahoo! Photos: High-quality 4x6 digital prints for 25�