Re: [PHP-DB] INSERT problem--Need help urgently

Hi Jason: okay I have echoed $sql and before I did enter anything into the form, the form is already inserting null values into the database. Maybe that is the reason for the duplicate row of null values. Do you know how can I prevent this from happening??Hope to hear from you soon. Regards,

Re: [PHP-DB] INSERT problem--Need help urgently

sorie it was my typo error. actually in my code there is already {} with my first if statement and there is also a space after VALUES... I still need some help here with my code...how can I prevent it from inserting a duplicate row of null values>??? Anyone can help ? Thanks in advance. Regard

[PHP-DB] INSERT problem--Need help urgently

Hi all: I was having some problem with INSERT query here and do hope to get some help real soon. I have a form whereby once the form is submitted, the values entered shall be inserted into the database. Next I have also written a SELECT query to view the records which have just been inserted i

[PHP-DB] Select Query--Need help urgently

Hi all, I am having having a slight problem with SELECT query statement. I have a table "Class". The table are as follow: ++--+--+-+-++ | Field | Type | Null | Key | Default | Extra | ++---

[PHP-DB] INSERT dynamic values problem - Need help urgently!

Hi all: I am trying to create a registration form whereby one of the user input is as follows: (a drop down menu with values retrieved from DB): - Class Code: " .$row["class_code"]. ""; } $result = $db->query($sql);

[PHP-DB] INSERT into mysql from dynamic drop down

Hi all, Right now i would like to INSERT the values from a dynamic drop down menu into mysql database but encountered some problem here. Values in the drop down menu are retrieved from DB as follows: " .$row["tutor_name"]. ""; } $result = $db->query($sql); ?> -

[PHP-DB] Problem with INSERT Query

Hi all: Well, when i bring out the page with the drop down list it was able to display all tutors' names from "tutor_name" column. Anyway here's a review of my code (snip) again before i continue: --- $sql = "INSERT I

RE: [PHP-DB] Re: Display Mysql Result in drop down list

Hi: Well, when i bring out the page with the drop down list it was able to display all tutors' names from "tutor_name" column. Anyway here's a review of my code (snip) again before i continue: --- $sql = "INSERT INTO cla

[PHP-DB] INSERT into 2 different tables??

Hi all: Just wana ask is it possible to have only one form and when i submit, it should INSERT data into 2 diff tables?? Hope u get my idea... Merry X'Mas!! Irin. -- PHP Database Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

Re: [PHP-DB] Re: Display Mysql Result in drop down list

Hi: then now i'm trying to select a value from the drop down list, echo the value i have selected and lastly INSERT the value into DB with the following : Below's a snippet of code: //retrieve all "tutor_name" from DB and display in drop down list " .$row ["tutor_name"]. ""; } $result = $db

[PHP-DB] Display Mysql Result in drop down list

Hi all, Right now I'm trying to retrieve one of the column "tutor_name" and display al the tutor's name in a drop down list. The problem now is, the drop down list only manage to display 1 record from that row instead of all tutor's name under "tutor_name" column...wonder where the problem li

[PHP-DB] Thanks alot Griffiths!=)

Thanks alot for your help! That was very careless of me..;p I was stil quite a newbie to php-db coding... By the way one last favour to ask of u hope u dun mind... I have a script which was used to INSERT data into the database: (See script below): But everytime when I execute it I got the er

RE: [PHP-DB] Need help!!!

Yes I'm using a hyperlink to access this page (see below) echo "Modify"; and so I defined my variable "$tutor_id" as $tutor_id = $_GET['tutor_id']; but i still got the error: "Undefined index: tutor_id in /usr/local/apache/htdocs/tutor/edit.php " WHat could be the problem?;(

[PHP-DB] Need help!!!

Hi.. Anyone can help me with this script???I kept getting error saying i did not define the variable "$tutor_id" and whenever I define it as $tutor_id = $_POST ['tutor_id'], I got error saying there was an "Undefined index"! What was exactly the problem?? Need help desperately , greatly app

[PHP-DB] Need HELP with UPDATE & DELETE

Hi all, Right now I'm trying to do a UPDATE and DELETE queries using the following code below: But i kept getting error "Undefined variable: tutor_id in /usr/local/apache/htdocs/tutor/edit.php on line . Didn't I defined my variable "$tutor_id"I thought I have... Anyone can help??? Tha