SQL SERVER – Ranking Functions – RANK( ), DENSE_RANK( ), and ROW_NUMBER( ) – Day 12 of 35

In this blog post we will discuss about Ranking Functions like RANK( ), DENSE_RANK( ), and ROW_NUMBER( ).

Ranking Functions (Part 1)

There are four ranking functions in SQL server. Today we will look at RANK( ), DENSE_RANK( ), and ROW_NUMBER( ).These functions all have the same basic behavior. Where they differ is in the handling of tie values. These three functions produce identical results, until a tying value in your data is present.

Until the tie data is encountered, the results are identical (see Josh and Kevin’s score of 9.6. We will explain more on this later. Let’s look at RANK( ) using a Grant table example sorted by largest to smallest grant amount (descending sort order).

A smart way to begin your ranking functions is to first write a SELECT statement with an ORDER BY clause. Every ranking function retrieves your sort order from the ORDER BY clause. In this case, the item is grant amount arranged in descending order. Add RANK( ) to the SELECT list. Remember that all ranking functions need an OVER( ) clause. If you then SQL Server reminds you as you can see from the error message below.

All ranking functions need an OVER( ) clause and SQL Server provides another syntax clue for the ORDER BY clause. Ranking functions need the ORDER BY information to appear as an argument of the OVER( ) clause Move the ORDER BY clause inside the parentheses of OVER( ) clause.

Now we can look at how RANK() works with ties. The Grant table ranked over the Amount column, showing 2 pairs of tied amounts. There is a 2 way tie for 4th place so both the grants of $21,000 get a rank of 4. After that there is no 5^th place grant. There is a 6^th place grant so after ties you get a gap to match the numbered order. Observe that no ORDER BY clause follows the FROM clause, since we moved it in into the OVER( ) clause. Examine what the RANK( ) function has done here. We see straightforward rankings of 1, 2, and 3 for the first three unique amount values. Notice the tie for fourth place. Then there’s another tie for the next place, which is ranked as 6. After that we get to 8^th place as seen in the figure below.

How about Dense_Rank? After the 2 way tie for 4^th place the next value will get 5^th place. There will be no gaps after the tie with the DENSE_RANK(). DENSE_RANK( ) is very useful for finding the five highest distinct amounts, because it finds each amount and has no gap in its numbering sequence.

Note: If you want to setup the sample JProCo database on your system you can watch this video. For this post you will want to run the SQLQueriesChapter7.0Setup.sql script from Volume 2.

Question 12

The figure below shows the scores of 6 contest winners. Tom is the highest
and Eric made 6^th place. There were 6 people but only 5 distinct scores.

You are writing a query to list the 5 highest distinct scores. The Ranked field should be called ScoreRating. You have written the following code.

SELECT * FROM
(SELECT  *  More code here.
FROM [Contestants])  AS ContestantFinal
WHERE ScoreRating <= 5

What code will achieve this goal?

1. SUM(*) OVER(ORDER BY Score DESC) as ScoreRating
2. RANK( ) OVER(ORDER BY Score DESC) as ScoreRating
3. COUNT(*) OVER(ORDER BY Score DESC) as ScoreRating
4. DENSE_RANK( ) OVER(ORDER BY Score DESC) as ScoreRating

Rules:

Please leave your answer in comment section below with correct option, explanation and your country of resident.
Every day one winner will be announced from United States.
Every day one winner will be announced from India.
A valid answer must contain country of residence of answerer.
Please check my facebook page for winners name and correct answer.
Winner from United States will get Joes 2 Pros Volume 2.
The contest is open till next blog post shows up at which is next day GTM+2.5.

Reference: Pinal Dave (https://blog.sqlauthority.com)