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Article ID: 186133 - Last Review: September 17, 2011 - Revision: 7.0

This article was previously published under Q186133

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SUMMARY

This article describes how to dynamically rank rows when you perform a SELECT statement by using a flexible method, which may be the only possible solution and which is faster than the procedural solution. Row numbering or ranking is a typical procedural issue. The solutions are typically based on loops and temporary tables; therefore, they are based on SQL Server loops and cursors. This technique is based on an auto join. The chosen relationship is typically "is greater than." Count how many times each element of a particular set of data fulfills the relationship "is greater than" when the set is compared to itself.

Note The following examples are based on the pubs database. By default, the Northwind sample database and the pubs sample database are not installed in SQL Server 2005. These databases can be downloaded from the Microsoft Download Center. For more information, visit the following Microsoft Web site:
http://go.microsoft.com/fwlink/?linkid=30196 (http://go.microsoft.com/fwlink/?linkid=30196)
After you download SQL2000SampleDb.msi, extract the sample database scripts by double-clicking SQL2000SampleDb.msi. By default, SQL2000SampleDb.msi will extract the database scripts and a readme file into the following folder:
C:\SQL Server 2000 Sample Databases
Follow the instructions in the readme file to run the installation scripts.

If you are using SQL Server 2005

We recommend that you use ranking functions that are provided as a new feature in SQL Server 2005. For more information about the ranking functions, visit the following Microsoft Developer Network (MSDN) Web site:
http://msdn2.microsoft.com/en-us/library/ms189798.aspx (http://msdn2.microsoft.com/en-us/library/ms189798.aspx)

Example 1

In this example:
  • Set 1 is authors.
  • Set 2 is authors.
  • The relationship is "last and first names are greater than."
  • You can avoid the duplicate problem by comparing the first + last names to the other first + last names.
  • Count the number of times the relationship is fulfilled by count(*).
Query:
   select rank=count(*), a1.au_lname, a1.au_fname
   from authors a1, authors a2
   where a1.au_lname + a1.au_fname >= a2.au_lname + a2.au_fname
   group by a1.au_lname, a1.au_fname
   order by rank
				
Use the following code in SQL Server 2005.
   select rank() OVER (ORDER BY a.au_lname, a.au_fname) as rank, a.au_lname, a.au_fname
   from authors a
   order by rank 
Result:
   Rank        Au_Lname              Au_Fname
   ----        --------------        -----------
     1         Bennet                Abraham
     2         Blotchet-Halls        Reginald
     3         Carson                Cheryl
     4         DeFrance              Michel
     5         del Castillo          Innes
     6         Dull                  Ann
     7         Greene                Morningstar
     8         Green                 Marjorie
     9         Gringlesby            Burt
    10         Hunter                Sheryl
    11         Karsen                Livia
    12         Locksley              Charlene
    13         MacFeather            Stearns
    14         McBadden              Heather
    15         O'Leary               Michael
    16         Panteley              Sylvia
    17         Ringer                Albert
    18         Ringer                Anne
    19         Smith                 Meander
    20         Straight              Dean
    21         Stringer              Dirk
    22         White                 Johnson
    23         Yokomoto              Akiko

   (23 row(s) affected)
				

Example 2

In this example:
  • Rank stores by the number of books sold.
  • Set 1 is the number of books sold by store: select stor_id, qty=sum(qty) from sales group by stor_id.
  • Set 2 is the number of books sold by store: select stor_id, qty=sum(qty) from sales group by stor_id.
  • The relationship is "the number of books is greater than."
  • To avoid duplicates, you can (as an example) compare price*qty instead of qty.
Query:
   select rank=count(*), s1.stor_id, qty=sum(s1.qty)
   from (select stor_id, qty=sum(qty) from sales group by stor_id) s1,
        (select stor_id, qty=sum(qty) from sales group by stor_id) s2
   where s1.qty >= s2.qty
   group by s1.stor_id
   order by rank
				
Result:
   Rank     Stor_Id    Qty
   ----     -------    ---
   1         6380        8
   2         7896      120
   3         8042      240
   4         7067      360
   5         7066      625
   6         7131      780

   (6 row(s) affected)
				
Note The values in the Qty column are incorrect. However, the ranking of stores based on the quantity of books sold is correct. This is a defect of this method. You can use this method to return the ranking of stores if you do not care about the wrong quantity in the result.

Use the following code in SQL Server 2005.
select row_number() over (order by qty desc) as rank,s1.stor_id,s1.qty
from (select stor_id, qty=sum(qty) from sales group by stor_id) as s1
Result:
rank     stor_id  qty
-------  -------  ------
1        7131     130
2        7066     125
3        7067     90
4        8042     80
5        7896     60
6        6380     8

(6 row(s) affected)
Note In SQL Server 2005, you can receive the correct result of the ranking and the quantity when you use the ranking functions.

Example 3

In this example:
  • Rank the publishers by their earnings.
  • Set 1 is the total sales by publisher:
            select t.pub_id, sales=sum(s.qty*t.price)
            from sales s, titles t
            where s.title_id=t.title_id
              and t.price is not null
            group by t.pub_id
    					
  • Set 2 is the total sales by publisher:
            select t.pub_id, sales=sum(s.qty*t.price)
            from sales s, titles t
            where s.title_id=t.title_id
              and t.price is not null
            group by t.pub_id
    					
  • The relationship is "earns more money than."
Query:
   select rank=count(*), s1.pub_id, sales=sum(s1.sales)
   from    (select t.pub_id, sales=sum(s.qty*t.price)
           from sales s, titles t
           where s.title_id=t.title_id
             and t.price is not null
           group by t.pub_id) s1,
           (select t.pub_id, sales=sum(s.qty*t.price)
           from sales s, titles t
           where s.title_id=t.title_id
             and t.price is not null
           group by t.pub_id) s2
   where s1.sales>= s2.sales
   group by s1.pub_id
   order by rank
				
Result:
   Rank     Pub_Id   Sales
   ----     ------   --------
   1         0736    1,961.85
   2         0877    4,256.20
   3         1389    7,760.85

   (3 row(s) affected)
				
Note The values in the Sales column are incorrect. However, the ranking of publishers based on the earnings is correct.

Use the following code in SQL Server 2005.
select rank() over (order by sales desc) as rank,s1.pub_id,s1.sales 
from (select t.pub_id, sales=sum(s.qty*t.price)
     from sales s inner join titles t
     on s.title_id=t.title_id
     where  t.price is not null
     group by t.pub_id) as s1
Result:
rank     pub_id  sales
-------  ------  ---------
1        1389    2586.95
2        0877    2128.10
3        0736    1961.85

(3 row(s) affected)
				
Note You receive the correct result of the ranking and the earning when you use the ranking functions.

Drawbacks

  • Because of the cross join, this is not designed for working with a large number of rows. It works well for hundreds of rows. On large tables, make sure to use an index to avoid large scans.
  • This does not work well with duplicate values. When you compare duplicate values, discontinuous row numbering occurs. If this is not the behavior that you want, you can avoid it by hiding the rank column when you insert the result in a spreadsheet; use the spreadsheet numbering instead.

    Note If you are using SQL Server 2005, you can use the row_number() function to return the sequential number of a row, regardless of the duplicate rows.
Example:
   select rank=count(*), s1.title_id, qty=sum(s1.qty)
   from (select title_id, qty=sum(qty) from sales group by title_id) s1,
        (select title_id, qty=sum(qty) from sales group by title_id) s2
   where s1.qty >= s2.qty
   group by s1.title_id
   order by rank
Result:
   Rank    Title_Id    Qty
   ----    --------    ----
   1       MC2222        10
   4       BU1032        60
   4       BU7832        60
   4       PS3333        60
   7       PS1372       140
   7       TC4203       140
   7       TC7777       140
   10      BU1111       250
   10      PS2106       250
   10      PS7777       250
   11      PC1035       330
   12      BU2075       420
   14      MC3021       560
   14      TC3218       560
   15      PC8888       750
   16      PS2091      1728

   (16 row(s) affected)
				

Benefits

  • You can use these queries in views and result formatting.
  • You can shift the lower-ranked data more to the right.
Example 1:
   CREATE VIEW v_pub_rank
   AS
   select rank=count(*), s1.title_id, qty=sum(s1.qty)
   from (select title_id, qty=sum(qty) from sales group by title_id) s1,
        (select title_id, qty=sum(qty) from sales group by title_id) s2
   where s1.qty >= s2.qty
   group by s1.title_id
				
Query:
   select  publisher=convert(varchar(20),replicate (' ', power(2,rank)) +
           pub_id +
           replicate(' ', 15-power(2,rank))+': '),
           earnings=qty
   from v_pub_rank
				
Result:
   Publisher       Earnings
   -------------   --------
     0736          : 1,961.85
       0877        : 4,256.20
           1389    : 7,760.85
				
Use the following code in SQL Server 2005.
CREATE VIEW v_pub_rank
AS
select rank() over (order by sales) as rank,s1.pub_id,s1.sales 
	from (select t.pub_id, sales=sum(s.qty*t.price)
	from sales s, titles t
	where s.title_id=t.title_id
	and t.price is not null
	group by t.pub_id) as s1
GO

select  publisher=convert(varchar(20),replicate (' ', power(2,rank)) +
	pub_id + replicate(' ', 15-power(2,rank))+': '),
	earnings=sales
from v_pub_rank order by rank
GO
Result:
publisher            earnings
-------------------- ---------------------
  0736             : 1961.85
    0877           : 2128.10
        1389       : 2586.95

(3 row(s) affected)

Example 2:
   CREATE VIEW v_title_rank
   AS
   select rank=count(*), s1.title_id, qty=sum(s1.qty)
   from (select title_id, qty=sum(qty) from sales group by title_id) s1,
        (select title_id, qty=sum(qty) from sales group by title_id) s2
   where s1.qty >= s2.qty
   group by s1.title_id
				
Query:
   select  Book=convert(varchar(45),replicate (' ', 2*rank) +
           title_id +
           replicate(' ', 35-2*rank)+': '),
           qty
   from v_title_rank
   order by rank
				
Result:
   Book                                          Qty
   -------------------------------------------   ----
     MC2222                                 :      10
           BU1032                           :      60
           BU7832                           :      60
           PS3333                           :      60
                 PS1372                     :     140
                 TC4203                     :     140
                 TC7777                     :     140
                       BU1111               :     250
                       PS2106               :     250
                       PS7777               :     250
                         PC1035             :     330
                           BU2075           :     420
                               MC3021       :     560
                               TC3218       :     560
                                 PC8888     :     750
                                   PS2091   :    1728

   (16 row(s) affected)
				
Use the following code in SQL Server 2005.
CREATE VIEW v_title_rank
AS
select rank() over (order by qty) as rank, s1.title_id,s1.qty
from (select title_id, qty=sum(qty) from sales group by title_id) as s1
GO

select Book=convert(varchar(45),replicate (' ', 2*rank) +
title_id + replicate(' ', 35-2*rank)+': '), qty
from v_title_rank
order by rank
GO
Result:
Book                                          qty
--------------------------------------------- -----------
  MC2222                                 :    10
    BU1032                               :    15
    BU7832                               :    15
    PS3333                               :    15
          TC4203                         :    20
          TC7777                         :    20
          PS1372                         :    20
                BU1111                   :    25
                PS7777                   :    25
                PS2106                   :    25
                      PC1035             :    30
                        BU2075           :    35
                          MC3021         :    40
                          TC3218         :    40
                              PC8888     :    50
                                PS2091   :    108

(16 row(s) affected)

 

APPLIES TO
  • Microsoft SQL Server 2000 Standard Edition
  • Microsoft SQL Server 4.21a Standard Edition
  • Microsoft SQL Server 6.0 Standard Edition
  • Microsoft SQL Server 6.5 Standard Edition
  • Microsoft SQL Server 7.0 Standard Edition
  • Microsoft SQL Server 2005 Standard Edition
  • Microsoft SQL Server 2005 Developer Edition
  • Microsoft SQL Server 2005 Enterprise Edition
  • Microsoft SQL Server 2005 Express Edition
  • Microsoft SQL Server 2005 Workgroup Edition
Keywords: 
kbsqlsetup kbhowtomaster KB186133
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