Question

mysql category query code

Asked by: killer455

- listing and category is a many to many relationship meaning listings
can have multiple categories, and categories can have multiple listings
and this is why listing2category is used
- feel free to change indexes if you need to but give reasons why
- below are sample queries needed

CREATE TABLE `item2cat` (
  `item_id` int(11) unsigned NOT NULL default '0',
  `cat_id` int(11) unsigned NOT NULL default '0',
  PRIMARY KEY  (`item_id`,`cat_id`),
  KEY `cat_id_2` (`cat_id`,`item_id`)
) TYPE=MyISAM;

CREATE TABLE `category` (
  `cat_id` mediumint(8) unsigned NOT NULL default '0',
  `title` varchar(255) NOT NULL default '',
  `description` text NOT NULL,
  `p` mediumint(8) unsigned default NULL,
  PRIMARY KEY  (`cat_id`),
  KEY `p` (`p`)
) TYPE=MyISAM;

CREATE TABLE `item` (
  `item_id` mediumint(8) unsigned NOT NULL auto_increment,
  `approved` tinyint(1) NOT NULL default '1',
  `title` varchar(100) NOT NULL default '',
  `membership` tinyint(2) NOT NULL default '1',
  `expires` datetime NOT NULL default '0000-00-00 00:00:00',
  PRIMARY KEY  (`item_id`),
  KEY `membership` (`membership`),
  KEY `date_expire` (`date_expire`,`admin_approved`),
  FULLTEXT KEY `firmname` (`title`)
) TYPE=MyISAM;

- Select all items in a given category that have not expired (expires <NOW()) and are approved (approved=1)
  and order by the membership type (higher memberships show first) then limit in order to page results

- Get a result set containing 2 columns, cat_id, and item_count.  item_count is a sum of the items in cat_id and all
  of cat_id's children.

- We want to display 2 levels of the category on a single page at a time showing the main category name
  followed by its children.  Beside both the category name and the children name a number which is the item count for that category.
  So we need 1 or 2 queries that can get the information needed to display this.   Tedious/slow PHP code should not be required to
  display the tree from the result set.  We would need to set the root cat_id (NULL for top level) and the query could get the corresponding data.

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Asked On
2006-11-16 at 13:23:00ID22063613
Tags

category

,

listing2category

Topic

MySQL Server

Participating Experts
1
Points
250
Comments
5

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Answers

 

by: Raynard7Posted on 2006-11-16 at 13:35:38ID: 17960317

1.

Select * from item i, item2cat i2 where i.item_id = i2.item_id and i.expires < now() and i.approved = 1 and i2.cat_id = 3

where 3 is the cat you want to choose from.

2.  What do you mean by category's children? how is this defined? similarly with q 3 this does not seem to have a parent child relationship.  If there is you can not do this with a query (unless you only have a fixed number of levels) you would have to do this with a stored procedure.

If you just wanted a count of items you would do

select count(*) from item2cat where cat_id = 3

Mysql has no default functions to find parents or children - so only if you are using mysql 5 would this work - you would have to have a function where you pass the category id to it and that of the parent you wish to check for - and it recursivley goes through all parents and checks if that parent category falls into the relationship and return true if it does.

Finally - you should only ask one question at a time on EE - and the mysql space is probably not the best place to ask PHP questions (q 3)

 

by: killer455Posted on 2006-11-16 at 15:19:49ID: 17961273

Select * from item i, item2cat i2 where i.item_id = i2.item_id and i.expires < now() and i.approved = 1 and i2.cat_id = 3
>> This does not sort by membership as mentioned

2.  What do you mean by category's children? how is this defined? similarly with q 3 this does not seem to have a parent child relationship.  If there is you can not do this with a query (unless you only have a fixed number of levels) you would have to do this with a stored procedure.
>> category table, cat_id and p, where p is the parent forms the relationship.  Ok so what if you do know the levels?

select count(*) from item2cat where cat_id = 3
>> This is just for one cat, it does not count children items

Mysql has no default functions to find parents or children - so only if you are using mysql 5 would this work - you would have to have a function where you pass the category id to it and that of the parent you wish to check for - and it recursivley goes through all parents and checks if that parent category falls into the relationship and return true if it does.
>> Sure but you can form the relationship through self joins?

Finally - you should only ask one question at a time on EE - and the mysql space is probably not the best place to ask PHP questions (q 3)
>> The question all involve the same database schema and relate directly to Mysql, I dont see how you consider that PHP?  I simply said no PHP code should be needed to process results.  

 

by: Raynard7Posted on 2006-11-16 at 15:28:19ID: 17961338

1.
Select * from item i, item2cat i2 where i.item_id = i2.item_id and i.expires < now() and i.approved = 1 and i2.cat_id = 3 order by membership

2. so p indicates a parent? so joining on where p from the table = cat_id?

3.  I did not include children as it was not clear what a child was.

I'll get back to you shortly (2 hurs or so)

I still beleive that if you are asking three different questions that can have three different answers and you want three different queries this is three different posts - especially from the help link where it makes this clear.

I apologise for the PHP comment - I assumed when you said that you do not want tedious or slow php code you wanted efficient and fast php code

20120131-EE-VQP-002

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