The scenario was like I had a customer apllication form for plots. Mostly getting the customer data and it has a droopdown field showing the available plots from the plots table where plots.clients_id='NULL' and status='Null'.

Now after the customer has selected a plot i want to store the client_id from the session variable into the plots.client_id (a foreign key from clients table) where plots.plot_id=selected plot id from the form.

The query i am trying to run does not store the value of the client id into the plots table. The query is as follows:

<?php
include("connect2db.php");
$plot_id=$_POST['plot_select'];
//echo $plot_id; //the value is printed
$client_id= $_SESSION['id'];
//echo $client_id;
$query="INSERT INTO plots(client_id) VALUES ($client_id) where plots.plot_id='$plot_id'";
$result=mysql_query($query);
?>


Please help...thanks is advance...
AZEEM

• When you compare against a null value it's done like this:
  
  Code:

  where plots.clients_id is NULL and status is Null

  not like this:
  
  Code:

  where plots.clients_id='NULL' and status='Null'

• When you run any mysql function (doing the connection, going to the database and the running the SQL) it's a good idea to test if the function succeeded or not using the or die command:
  
  Code:

  $result=mysql_query($query) or die( "Problem with sql: $query" );

• It would also be a good idea to print out $query and try running the SQL in your database by hand. If it fails then you can see immediately what's wrong.

i can tell you right now what's wrong, you won't have to test your SQL (although that ~is~ good advice -- always test outside of php first)

you have:the INSERT VALUES statement does not allow a WHERE clause

you probably want UPDATE, not INSERT

check da manual for the correct UPDATE syntax

Ooops - not sure how I missed that glaring mistake. Well spotted.

Hi guys !

Thanks a lot for your solution...
The problem is resolved...

Thanks
Azeem