I'm trying to list my categories and how many items are in each
category. I'm not much on sub queries. How can I combine these two
items. I'd like to display the category and the corresponding number of
items in each
Cats:
Select
Category.Catego ryID,
Category.Catego ry,
Category.Sortin g
From
Category
Order By
Category.Sortin g Desc
Items:
Select Count(Listings. CategoryID)
From
Listings
Group By
Listings.Catego ryID
I've tried the following but can't get it to work:
Select
Category.Catego ryID,
Category.Catego ry,
Category.Sortin g,
(Select Count(Listings. CategoryID)
From
Listings WHERE Listings.Catego ryID = Category.Catego ryID) AS cnt From
Category
Order By
Category.Sortin g Desc 1  4603 
Select
Category.Catego ryID,
Category.Catego ry,
Category.Sortin g,
Count(Listings. CategoryID) as cnt From Category, Listing
WHERE Listings.Catego ryID = Category.Catego ryID)
group by
Category.Catego ryID,
Category.Catego ry,
Category.Sortin g
Order By
Category.Sortin g Desc
If you are going to continue doing SQL - a good reference book would be
in order. Also - the term (in this case) that you would use for Google
would be Joins, Cross Joins This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
by: knoak |
last post by:
Hi there,
I have a small question:
I have a table with lots of rows in it. Of course all have a different id,
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What i want is to do a count how much rows belong to a certain category.
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Hi,
I have a `category` table with columns cat_id and cat_name. And another
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Thanks~
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Web Applications (272)
Content Managers, Forum, E-Commerce, ...
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by: Barkster |
last post by:
I'm trying to list my categories and how many items are in each
category. I'm not much on sub queries. How can I combine these two
items. I'd like to display the category and the corresponding number of
items in each
Cats:
Select
Category.CategoryID,
Category.Category,
Category.Sorting
|
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Here is the structure of the 2 tables:
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<?xml version="1.0" encoding="utf-8"?>
<vol_issue vol_issue_id="200802">
...
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