Re: [PHP-DB] Help with a query

2003-03-06 Thread Corne' Cornelius
Why not use:

SELECT a.item_id, a.subtotal, a.quantity
  FROM shopping_cart a, orders b
 WHERE b.session_id = session_id()
   AND b.customer_id = $customer_id
   AND a.order_id = b.border_id
?

!Exclude Disclaimer!

Jonathan Villa wrote:

I can't figure this query out.

I want to pull data from 2 tables but using a primary key from one.

Here is what I think it should look like.

SELECT item_id, subtotal, quantity from shopping_cart WHERE order_id =
(SELECT order_id FROM orders WHERE session_id = session_id()) AND
customer_id = $customer_id;

-Jonathan








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Re: [PHP-DB] Storing images in MySQL table

2003-02-19 Thread Corne' Cornelius
Milan,

The only disadvantage i've had of storing images in DB instead of 
Filesystem, is that when you use a PHP script to output the image to a 
client browser, MSIE doesn't always accept a suggested filename so it 
might try and save it as your-script.php?img=2.

Other then that, keeping images in DB is a lot easier since you don't 
have to keep your DB in sync with Filesystems ie. Deleted records in DB, 
but images still exist.


[EMAIL PROTECTED] wrote:

Hi everybody,
	i want to store some articles and images to them in MySQL 
db. could you just give advice if it is better to store the image in BLOB 
or if to save on server and in db just have it's URL.


			Hi Milan


 





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Re: [PHP-DB] More help with mysql

2003-02-18 Thread Corne' Cornelius
Yo Evan,

Have you got an '@' in front of the mysql_query() to supress errors ?



Evan Morris wrote:


Hi all

I have successfully managed to connect to the mysql database, using the
following code:

mysql_connect('localhost','php','***') || die (Unable to connect to
MySQL server.);
$db = mysql_select_db(DB_NAME) || die (Unable to select requested
database.);

I then issue the following command:

$result = mysql_query(SELECT * FROM SOME_TABLE) || die(mysql_error());

This does not result in any error.

Now, of course, I want to work with the data returned, so I go:

if (mysql_num_rows($result)  0)
   {
   // do some stuff
   }
else
   {
   // do some other stuff
   };

However, this gives me: Supplied argument is not a valid MySQL result
resource.

If I go: echo $result, I get '1', so I know $result actually has some
value.

What am I doing wrong? Basically, I want to do the simplest thing: get data
out of a table and step through the results, displaying them one by one. Why
is this hard?

Evan Morris
[EMAIL PROTECTED]
Tel: +27 11 792 2777
Fax: +27 11 792 2711
Cell: +27 82 926 3630



 





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Re: [PHP-DB] Multiple select box using mysql

2003-02-18 Thread Corne' Cornelius
You have to give the Mutli select box a name as an array. eg:

SELECT NAME=experts[] MULTIPLE SIZE=5
OPTION VALUE=key1val1/OPTION
/SELECT

then in PHP, you can access $experts as an array.

while (list ($key, $val) = each($experts))  {
  print $key = $valBR\n;
}


Jason End wrote:


I writing a software catalog that features experts,
who a people especially skilled in a piece of
software. My db, has a software table and an
experts table. 
I need to have a page that generates 2 multiple select
boxes, each on with the data from each table. 
From there one will be able to choose the software

program and the experts, which will then record the
software-expert relationship in the DB. 

The problem is that I don't really understand how to
populate multi select boxes with db data, and then
have the selections passed back as inserts. 

Can someone clue me in?

thanks, 

Jay

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