Re: [PHP-DB] Creating drop down lists
Your select is only selecting the field 'root' but then you're trying to get
the field 'username' in your while loop. So, you can add username to your
select (or change it to *):
select root, username from training
select * from training
Regards,
Bruce
>>> Scott <[EMAIL PROTECTED]> 11/03/2007 12:33:37 a.m. >>>
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:
$query = mysql_query("SELECT root FROM training");
echo "";
while ($r = mysql_fetch_array($query))
{
$user = $r["username"];
echo "$user";
}
echo "";
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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RE: [PHP-DB] Creating drop down lists
close
$query = mysql_query("SELECT root FROM training");
echo "";
if($mysql_num_rows($result)>0){
while ($r = mysql_fetch_array($query))
{
$user = $r["username"];
echo "$user";
}
}
echo "";
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:
$query = mysql_query("SELECT root FROM training");
echo "";
while ($r = mysql_fetch_array($query))
{
$user = $r["username"];
echo "$user";
}
echo "";
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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AOL IM: Jestrix1
--
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