Have tried for a day to figure out how to dynamically fill a second drop-down list from the first using PHP and Javascript. Still new to both would greatly appreciate some direction. Below is an example. Thanks in advance. :)
All of this is inside an HTML form so that the user can add a request for computer equipment.
1. Drop-down list one is of vendors that I've acquired from a php call to mysql to select all vendors in a table.
2. Want the second drop-down list to dynamically acquire from mysql the vendor contacts for the vendor chosen in drop-down list 1.
The rest of the form I can handle. Will PHP allow me to do this without Javascript? I know that I could just have a page list the drop-down list of the vendors inside a form and have submit button there. Then $_post that value in another page to fill the second list and display the remaining html to get the remaining data, but it would be nice if I could do both lists on one page. I've searched the web all day looking for examples of using php and mysql with Javascript to do this, but haven't found any that I can get a handle on. Any help would be greatly appreciated. :)
On Monday, August 18, 2003, at 07:11 AM, php-db-digest-help@lists.php.net wrote:
php-db Digest 18 Aug 2003 11:11:40 -0000 Issue 1987
Topics (messages 29926 through 29934):
Re: AGONIZING Mysql Select DB issue. 29926 by: Micah Stevens
test - Please ignore 29927 by: vish.kohli
Query runs fine on Console but not in PHP 29928 by: vish.kohli
mysql date select statement 29929 by: Wendell Frohwein 29930 by: John W. Holmes 29931 by: Shahmat Dahlan
Is there a better way to write this? 29932 by: Chris Payne
Popup menu problem 29933 by: Kim Kohen 29934 by: John W. Holmes
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----------------------------------------------------------------------
From: Micah Stevens <micah@raincross-tech.com>
Date: Sun Aug 17, 2003 4:56:24 PM US/Eastern
To: Thomas Deliduka <thomas@xenocast.com>, PHP-DB List <php-db@lists.php.net>
Subject: Re: AGONIZING Mysql Select DB issue.
Try using the SQL to select which database.
example, instead of:
select * from table1
use:
select * from database1.table1
if that works, and the php command doesn't that may mean the the mysql client
lib is broken, although, I've been using it with mysql 4 and it seems to work
fine.
-Micah
On Sunday 17 August 2003 1:49 pm, Thomas Deliduka wrote:I'm not making two connections, I'm making one and only one call to mysql_connect. Also, there is no way in that function as per the definition page of it (http://us3.php.net/mysql_connect) to have the database selected as per your example below.
With my connection though, when I do: $dbh = Mysql_connect(blah, blah, blah) Mysql_select_db("db1")
I do call: Mysql_query("query", $dbh);
For some reason even though I am calling mysql_select_db("db1") it is
latching onto the first available database it has access to (or not as the
case/permissions may be) and chooses "db2" instead.
I don't know why, connecting to the MySQL 3.23 it selects the right
database, connecting to the Mysql 4.x server it doesn't allow a selecting
of the table even though I do the select function and it returns true as it
was selected properly.
On 8/16/03 12:23 AM this was written:If you are doing this:
$dbh = mysql_connect("db1", blah blah blah); $dbh2 = mysql_connect("db2", blah blah blah);
Then
$r = mysql_query("select * from mytable");
will use the db2 connection, because it is the most recent. However, if
you do this:
$r = mysql_query("select * from mytable", $dbh);
it will use the first connection, as specified in the handle that is
passed back by mysql_connect. mysql_query uses the most recent
connection by default; you can override this action by specifying which
DB handle to use for a given query. Replace $dbh with $dbh2 to select
from tables on the second database.
Peter
On Fri, 15 Aug 2003, Thomas Deliduka wrote:Here's the stats:
Two servers:
Server 1, Mysql 4.0.12, PHP 4.3.2, apache 1.3.27 Server 2, Mysql 4.0.14, PHP 4.3.2, apache 1.3.27
--
Thomas Deliduka IT Manager ------------------------- Xenocast Street Smart Media Solutions http://www.xenocast.com/
From: "vish.kohli" <vish.kohli@attbi.com> Date: Sun Aug 17, 2003 4:50:23 PM US/Eastern To: php-db@lists.php.net Subject: test - Please ignore
Please ignore, this is a test
From: "vish.kohli" <vish.kohli@attbi.com> Date: Sun Aug 17, 2003 4:56:39 PM US/Eastern To: php-db@lists.php.net Subject: Query runs fine on Console but not in PHP
Hi,
I am trying to execute a stored procedure in PHP via mssql_query ()
function. I pass 8 parameters to it.
The query runs fine in SQL Query Analyzer but when I run it thru PHP, it
behaves strangely. I can get number of rows (mssql_num_rows()) and number of
fields (mssql_num_fields()) in PHP, but when I try to execute
mssql_fetch_object() or mssql_fetch_array() on the same result identifier, I
get a "Page could not be displayed" (standard internet error page).
Please help. Thanks in advance.
From: Wendell Frohwein <wendell@3rdpixeldesignstudio.com> Date: Mon Aug 18, 2003 12:07:43 AM US/Eastern To: php-db@lists.php.net Subject: mysql date select statement
Hello all, I have this page where I search and add up commissions in a mysql database. I want to select commissions between a certain date range. This is what im currently using but it does not seem to work.
$from="20030101"; $to="20031230";
$search=mysql_query("SELECT SUM(amount) FROM commissions WHERE date BETWEEN '$from' AND '$to'");
Thanks in advance for any help
wendell
From: "John W. Holmes" <holmes072000@charter.net> Date: Mon Aug 18, 2003 12:09:30 AM US/Eastern To: Wendell Frohwein <wendell@3rdpixeldesignstudio.com> Cc: php-db@lists.php.net Subject: Re: mysql date select statement Reply-To: holmes072000@charter.net
Wendell Frohwein wrote:
Hello all, I have this page where I search and add up commissions in a mysql database. I want to select commissions between a certain date range. This is what im currently using but it does not seem to work. $from="20030101"; $to="20031230"; $search=mysql_query("SELECT SUM(amount) FROM commissions WHERE date BETWEEN '$from' AND '$to'");
Always use mysql_error() with your queries to see the error. You do not have a GROUP BY clause in your query, so it's failing.
-- ---John Holmes...
Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/
PHP|Architect: A magazine for PHP Professionals – www.phparch.com
From: Shahmat Dahlan <shahmatd@sains.com.my> Date: Mon Aug 18, 2003 12:09:36 AM US/Eastern To: Wendell Frohwein <wendell@3rdpixeldesignstudio.com> Cc: php-db@lists.php.net Subject: Re: mysql date select statement
hav u tried using dashes to denote which portion is the year, month, or day ?
e.g. instead of 20030101, you'd probably want to use 2003-01-01.
Wendell Frohwein wrote:
Hello all, I have this page where I search and add up commissions in a mysql database. I want to select commissions between a certain date range. This is what im currently using but it does not seem to work. $from="20030101"; $to="20031230"; $search=mysql_query("SELECT SUM(amount) FROM commissions WHERE date BETWEEN '$from' AND '$to'"); Thanks in advance for any help wendell
From: Chris Payne <chris@planetoxygene.com> Date: Mon Aug 18, 2003 12:16:11 AM US/Eastern To: php <php-db@lists.php.net> Subject: Is there a better way to write this?
Hi there everyone,
I'm re-writing the messageboard that my boss uses, and I wanted to know if there's a cleaner way to do the below code. The first bit gets the name of the forum by the ID sent from the previous page to display the forum name, then the second bit gets some of the info for the message listing. The forum name is NOT stored in the actual messages table, it's in the forum index table I created.
The code I use is:
<?
include("../../connectionstart.php");
$query = "SELECT * FROM forums WHERE ForumID = '$ForumID' ORDER BY ForumID ASC";
$sql_result = mysql_query($query,$connection) or die("Couldn't execute query.");
while ($row = mysql_fetch_array($sql_result)) {
$Forum = $row["Forum"];
};
$query = "SELECT * FROM forummessages WHERE ForumID = '$ForumID' ORDER BY ForumID ASC";
$sql_result = mysql_query($query,$connection) or die("Couldn't execute query.");
while ($row = mysql_fetch_array($sql_result)) {
$id = $row["id"]; $MessageID = $row["MessageID"]; $ForumID = $row["ForumID"]; $subject = $row["subject"];
}; ?>
Thanks for taking a look :-)
Chris
From: Kim Kohen <kim@webguide.com.au> Date: Mon Aug 18, 2003 3:04:17 AM US/Eastern To: PHP <php-db@lists.php.net> Subject: Popup menu problem
G'day all,
I've struck a problem with an existing piece of code which I want to move
to a new one. I'm aware of the issues with register global being off as far
as forms are concerned, but I can't figure out why this snippet doesn't
work. It displays the code rather than the popups.
<?php $db_name = "torch"; $table_name = "stories"
$connection = @mysql_connect("localhost","user","paswword") or die("Couldn't
Connect.");
$db = @mysql_select_db($db_name, $connection) or die("Couldn't select
database.");
$getlist = mysql_query("SELECT distinct Writer FROM stories order by
Writer");
echo " <select name=\"WriterName\">\n";
echo " <option>\n";
while ($row = mysql_fetch_array($getlist)) {
echo ' <option value="'.$row["Writer"].'">'.$row["Writer"]."</option>\n";
}
echo " </select>\n";
?>
As I said, it works fine on our current web server (apache 1.3x with a
similar vintage PHP and MySQL) but I have built the new server with Apache
2.0.47, MySQL 4 and PHP 4.3.2 (on MacOSX 10.2.6). I've pretty much ruled
out MySQL because the query returns the correct result when run from the
CLI.
I have another couple of pages which use hard coded popups and they work
fine. PHP seems fine as I can display phpinfo() without problem.
Any help, as always, would be greatly appreciated.
cheers
kim
From: "John W. Holmes" <holmes072000@charter.net> Date: Mon Aug 18, 2003 7:11:32 AM US/Eastern To: Kim Kohen <kim@webguide.com.au> Cc: PHP <php-db@lists.php.net> Subject: Re: Popup menu problem Reply-To: holmes072000@charter.net
Kim Kohen wrote:
G'day all,
I've struck a problem with an existing piece of code which I want to move
to a new one. I'm aware of the issues with register global being off as far
as forms are concerned, but I can't figure out why this snippet doesn't
work. It displays the code rather than the popups.
If you're seeing the PHP source code in the window, then PHP is not configured correctly. Are you calling a file with a .php extension (dumb question, I know, but you never know).
There's no difference in calling a PHP file through a popup or through a regular link. The code you showed didn't look wrong, show us how you're calling this code.
-- ---John Holmes...
Amazon Wishlist: www.amazon.com/o/registry/3BEXC84AB3A5E/
PHP|Architect: A magazine for PHP Professionals – www.phparch.com
:) Gale L. Allen Jr Macintosh Support Specialist Phone: 919/412-5039 Homepage: http://homepage.mac.com/ateam3/Menu5.html iChat = ateam3@mac.com "Remember, It's only Rock 'n Roll, but I like it!" (:
