-Original Message-
From: Ashley Sheridan [mailto:a...@ashleysheridan.co.uk]
Sent: Friday, July 02, 2010 2:43 PM
To: b...@tottd.com
Cc: php-general@lists.php.net
Subject: Re: [PHP] SELECT AVG(rating)
On Fri, 2010-07-02 at 14:32 -0600, Ben Miller wrote:
> Hi - I have a MySQL table full of product reviews and I'm trying to select
> info for only the top 5 rated products. The only way I can figure out how
> to do it so far is something like:
>
> $query1 = mysql_query("SELECT * FROM products");
> for($i=1;$i<=mysql_num_rows($query1);$i++) {
> $row1 = mysql_fetch_array($query1,MYSQL_ASSOC);
> $query2 = mysql_query("SELECT AVG(rating) as rating FROM reviews
> WHERE product_id='" . $row1['product_id'] . "'");
> ...
> $product[$i]['name'] = $row1['product_name'];
> $product[$i]['rating'] = $row2['rating'];
> }
>
> And then use array functions to sort and display only the first 5.
>
> Is there any easier way to get this done with a single query - something
> like "SELECT AVG(rating) WHERE product_id=DISTINCT(product_id)"? <<= I
tried
> that - it didn't work. Would greatly appreciate any advice. Thanks,
>
> Ben
>
>
>
How about something like this (untested)
SELECT products.product_id, AVG(reviews.rating) AS rating
FROM products
LEFT JOIN reviews ON (reviews.product_id = products.product_id)
GROUP BY products.product_id
ORDER BY rating
LIMIT 1,5
I'm unsure about that order and limit there, so you might have to wrap
that inside of a temporary table query and take your 5 out of that. A
join is the right way to go with this though I reckon.
Thanks,
Ash
http://www.ashleysheridan.co.uk
Adding a DESC after ORDER BY rating did it perfectly. I had tried a few
JOIN queries, but was building them incorrectly. Thank you so much for your
help.
Ben
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
