Hi there.
Sorry to make a seperate post, but it is a seperate issue.

I have a page which starts off with a drop down list. This lists all customers that are stored in the Customers table and shows the cust_id. (Also, is it easy to show the cust_id as well as the name next to it?)

How do I code it so that when the user chooses a customer from the list, it will then display any records in the linked table (Timesheets) where the cust id is present.

So it will just show any timekeeping tasks that have been added against the chosen customer.

My code is as follows:

<h1>View Report</h1>
<p>Please choose a Customer ID from the drop down selection below, then press Submit to view a list of the tasks recorded against the select Customer.</p>
<?php
include 'config.php';
include 'opendb.php';
$res=mysql_query("SELECT * FROM customers order by cust_id") or die(mysql_error()); 
echo "<select name=myselect>"; 
while($row=mysql_fetch_assoc($res)) { 
 
echo "<option value=$row[ID]>$row[cust_id]</a></option>"; 
} 
echo "</select>"; 
include 'library/closedb.php';
?>

You may want to try the following. It will submit the information to the same page, and query the Timesheets table for the related data.

PHP Code:

<h1>View Report</h1>
<p>Please choose a Customer ID from the drop down selection below, then press Submit to view a list of the tasks recorded against the select Customer.</p>
<form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>">
<?php
include 'config.php';
include 'opendb.php';
$res=mysql_query("SELECT * FROM customers order by cust_id") or die(mysql_error()); 
echo "<select name=myselect>"; 
while($row=mysql_fetch_assoc($res)) { 
 
echo "<option value=$row[ID]>$row[cust_id]</a></option>"; 
} 
echo "</select>"; 
include 'library/closedb.php';
?>
<input type="submit" />
</form>
<?php
// check if form was submitted
if (isset($_POST['myselect']))
{
    $id = $_POST['myselect'];
    // query the records in Timesheets where the customer_id = $id
    $res = mysql_query("SELECT * FROM Timesheets WHERE customer_id=$id") or die(mysql_error());
    
    while($row = mysql_fetch_assoc($res)) 
    {
        // print here info about the current row
    }
}
?>

Hi there.
Thanks for the reply. I had added that code to my page, and when I run it I get the following error message:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1.

This is a copy of the code that I use, I had to change it due to table and row names being incorrect. To actually see it on the page i'm working on, go to http://www.lindendesign.co.uk/project/report1.php.

<h1>View Report</h1>
<p>Please choose a Customer ID from the drop down selection below, then press Submit to view a list of the tasks recorded against the select Customer.</p>
<form method="post" action="<?php echo $_SERVER['PHP_SELF'] ?>">
<?php
include 'config.php';
include 'opendb.php';
$res=mysql_query("SELECT * FROM customers order by cust_id") or die(mysql_error()); 
echo "<select name=myselect>"; 
while($row=mysql_fetch_assoc($res)) { 
 
echo "<option value=$row[ID]>$row[cust_id]</a></option>"; 
} 
echo "</select>"; 
include 'library/closedb.php';
?>
<input type="submit" />
</form>
<?php
// check if form was submitted
if (isset($_POST['myselect']))
{
    $id = $_POST['myselect'];
    // query the records in Timesheets where the customer_id = $id
    $res = mysql_query("SELECT * FROM timekeeping WHERE cust_id=$id") or die(mysql_error());
 
    while($row = mysql_fetch_assoc($res)) 
    {
        // print here info about the current row
    }
}
?>

There's a lost tag in the form. Change the line

PHP Code:

echo "<option value=$row[ID]>$row[cust_id]</a></option>"; 


to

PHP Code:

echo "<option value=$row[ID]>$row[cust_id]</option>"; 


I should work now.

hi,

thanks for the reply again,

i still have the same problem, even after removing the [/a].

Any ideas?

Thanks again,

Craig

Well, I saw your code and there's nothing in the value of the option tag.

PHP Code:

<option value=$row[ID]>$row[cust_id]</option> 


Looks like theres no ID in your database. Try changing to that:

PHP Code:

<option value=$row[cust_id]>$row[cust_id]</option>