What would be the most efficient way of creating my MySQL database - mysql

Question

I'm thinking of doing something like...
Table Pants (20 entrees)
| ID | Item | Description | Price
Table Shirts (20 entrees)
| ID | Item | Description | Price
Table Socks (5 entrees)
| ID | Item | Description | Price
Then I run a php foreach ID in $table, to populate the list on a page.
So would this be an efficient way of doing this since some of my tables might not have many entrees? Or should I have all the entrees in one table and add an extra field for the different categories and find another way of populating my page via the new field? Or maybe a greater idea?
Thanks for your time.

Answer 1

your model is not flexible and not efficient. Your second solution is the good one create a table named items and add an extra field named categories which should be a foreign key.
you can check this http://sqlfiddle.com/#!2/74c91/1, here is all you need
here is the code needed for the database creation :
/*!40101 SET #OLD_CHARACTER_SET_CLIENT=##CHARACTER_SET_CLIENT */;
/*!40101 SET NAMES utf8 */;
/*!40014 SET FOREIGN_KEY_CHECKS=0 */;
-- Dumping structure for table database.categories
CREATE TABLE IF NOT EXISTS `categories` (
`id` int(10) NOT NULL AUTO_INCREMENT,
`name` varchar(255) NOT NULL,
`created` datetime DEFAULT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- Dumping data for table database.categories: ~0 rows (approximately)
/*!40000 ALTER TABLE `categories` DISABLE KEYS */;
INSERT INTO `categories` (`id`, `name`, `created`) VALUES
(1, 'Pants ', '2012-12-15 11:37:27'),
(2, 'Shirts ', '2012-12-15 11:37:36'),
(3, 'Socks ', '2012-12-15 10:38:47');
/*!40000 ALTER TABLE `categories` ENABLE KEYS */;
-- Dumping structure for table database.clothes
CREATE TABLE IF NOT EXISTS `clothes` (
`id` int(10) NOT NULL AUTO_INCREMENT,
`item` varchar(255) NOT NULL,
`description` text,
`category_id` int(10) NOT NULL,
`created` datetime DEFAULT NULL,
PRIMARY KEY (`id`),
KEY `category_id` (`category_id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1;
-- Dumping data for table database.clothes: ~0 rows (approximately)
/*!40000 ALTER TABLE `clothes` DISABLE KEYS */;
INSERT INTO `clothes` (`id`, `item`, `description`, `category_id`, `created`) VALUES
(1, 'red socks', 'Red Socks Rocks', 3, '2012-12-15 10:39:06'),
(2, 'blue socks', 'nice socks too', 3, '2012-12-15 10:39:18'),
(3, 'grey pants', 'pretty cool', 1, '2012-12-15 10:40:34'),
(4, 'Blue Shirt', 'Nice blue shirt', 2, '2012-12-15 10:40:10');
/*!40000 ALTER TABLE `clothes` ENABLE KEYS */;
/*!40014 SET FOREIGN_KEY_CHECKS=1 */;
/*!40101 SET CHARACTER_SET_CLIENT=#OLD_CHARACTER_SET_CLIENT */;
And here is the code to retrieve your data:
SELECT
clothes.id AS clotheId,
clothes.item AS clotheItem,
clothes.description AS clotheDescription,
categories.id AS categoryId,
categories.name AS categoryName
FROM clothes
LEFT JOIN categories
ON clothes.category_id=categories.id;

Answer 2

I'd go with the latter.
ID | Item | Description | Price | Type
Then use a query like:
SELECT * FROM items WHERE type='$type'

Answer 3

I would not store this data in separate tables. Create one table, then include an indicator as to the type of item it is.
Similar to this:
create table ClothingItems
(
id,
typeid,
item,
description,
price
);
create table clothingType
(
typeid,
name
);
Then to query you can use:
select *
from clothingitems i
left join clothingType t
on t.typeid = t.typeid
where t.name = 'socks'

Answer 4

Well, it might not have so many entrees for the moment. But maybe in the future? Don't put yourself in a corner :).
Ask yourself the question; -Can an article belong to more than one category?
If NO, you should create at least 2 tables (one-to-many relation):
categories
id|title|description|...
articles
id|category_id|title|price|...
If YES, you should create at least 3 tables (many-to-many relation):
categories
id|title|description|...
articles
id|category_id|title|price|...
articles_categories
article_id|category_id
Best of luck :)