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RobertK · 01-10-2008, 03:22 PM

From what I see, your tables are set wrongly for rating tutorials. What you'd want is a new table like this:

sql Code:

CREATE TABLE `ratings`
(
  tut_id INT UNSIGNED NOT NULL,
  rating  TINYINT UNSIGNED NOT NULL
);

Then you'd insert the rating:

sql Code:

INSERT INTO `ratings` VALUES ($get, $rate);

You can get a rating by a subquery, like so:

sql Code:

SELECT 
  *,
  (
    SELECT
      AVG(`rating`)
    FROM
      `ratings` 
    WHERE
      `ratings`.`tut_id` = `tutorials`.`id`
  )
AS
  `rating`
FROM
  `tutorialz`;

Understand that I haven't tested this so I might have made a mistake, but this should work for you.

01-10-2008, 03:22 PM	#2 (permalink)
RobertK The Addict Join Date: Jan 2008 Location: USA Posts: 217 Thanks: 16	From what I see, your tables are set wrongly for rating tutorials. What you'd want is a new table like this: sql Code: CREATE TABLE `ratings` ( tut_id INT UNSIGNED NOT NULL, rating TINYINT UNSIGNED NOT NULL ); Then you'd insert the rating: sql Code: INSERT INTO `ratings` VALUES ($get, $rate); You can get a rating by a subquery, like so: sql Code: SELECT , ( SELECT AVG(`rating`) FROM `ratings` WHERE `ratings`.`tut_id` = `tutorials`.`id` ) AS `rating` FROM `tutorialz`; Understand that I haven't tested this so I might have made a mistake, but this should work for you. __________________ Programmers are in a race with the Universe to create bigger and better idiot-proof programs, while the Universe is trying to create bigger and better idiots. So far the Universe is winning.* - Rich Cook