Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
835 views
in Technique[技术] by (71.8m points)

sql server - Complicated SQL Query--finding items matching multiple different foreign keys

So imagine that you have a table of Products (ID int, Name nvarchar(200)), and two other tables, ProductsCategories (ProductID int, CategoryID int) and InvoiceProducts (InvoiceID int, ProductID int).

I need to write a query to produce a set of products that match a given set of invoice ids and category ids such that the list of products match all the specified categories and all the specified invoices, without falling back to dynamic SQL. Imagine I need to find a list of products that are in both categories 1 and 2 and in invoices 3 and 4.

As a start, I've written a stored-procedure that accept the category ids and invoice ids as strings, and parse them into tables:

 CREATE PROCEDURE dbo.SearchProducts (@categories varchar(max), @invoices varchar(max))
 AS BEGIN
      with catids as (select cast([value] as int) from dbo.split(@categories, ' ')),
           invoiceids as (select cast([value] as int) from dbo.split(@invoices, ' '))
           select * from products --- insert awesomeness here
 END

The different solutions I've come up with look awful, and perform worse. The best thing I've found is to generate a view comprised of left joins of all the criteria, but that seems very expensive and doesn't solve the issue of matching all of the different keys specified.


Update: This is an example query I wrote that yields the expected results. Am I missing any optimization opportunities? Like magical unicorn matrix operations by ninjas?

with catids as (select distinct cast([value] as int) [value] from dbo.split(@categories, ' ')),
  invoiceids as (select distinct cast([value] as int) [value] from dbo.split(@invoices, ' '))

  select pc.ProductID from ProductsCategories pc (nolock)
    inner join catids c on c.value = pc.CategoryID 
    group by pc.ProductID 
    having COUNT(*) = (select COUNT(*) from catids)  
  intersect
  select ip.ProductID from InvoiceProducts ip (nolock)
    inner join invoiceids i on i.value = ip.InvoiceID 
    group by ip.ProductID 
    having COUNT(*) = (select COUNT(*) from invoiceids)   
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Provided that you have unique indices on both (ProductID, CategoryID) and (ProductID, InvoiceID):

SELECT  ProductID
FROM    (
        SELECT  ProductID
        FROM    ProductInvoice
        WHERE   InvoiceID IN (1, 2)
        UNION ALL
        SELECT  ProductID
        FROM    ProductCategory pc
        WHERE   CategoryID IN (3, 4)
        ) q
GROUP BY
        ProductID
HAVING  COUNT(*) = 4

or, if your values are passed in CSV strings:

WITH    catids(value) AS
        (
        SELECT  DISTINCT CAST([value] AS INT)
        FROM    dbo.split(@categories, ' '))
        ), 
        (
        SELECT  DISTINCT CAST([value] AS INT)
        FROM    dbo.split(@invoices, ' '))
        )
SELECT  ProductID
FROM    (
        SELECT  ProductID
        FROM    ProductInvoice
        WHERE   InvoiceID IN
                (
                SELECT  value
                FROM    invoiceids
                )
        UNION ALL
        SELECT  ProductID
        FROM    ProductCategory pc
        WHERE   CategoryID IN
                (
                SELECT  value
                FROM    catids
                )
        ) q
GROUP BY
        ProductID
HAVING  COUNT(*) = 
        (
        SELECT  COUNT(*)
        FROM    catids
        ) + 
        (
        SELECT  COUNT(*)
        FROM    invoiceids
        )

Note that in SQL Server 2008 you can pass table-valued parameters into the stored procedures.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...