SQL SERVER – Tips from the SQL Joes 2 Pros Development Series – Advanced Aggregates with the Over Clause – Day 11 of 35

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Partitioning with the Over Clause (Part 2)

Yesterday we learned how the over clause can be used to compare your number against the overall aggregated number for an entire result set. Sometimes you might want your number to be compared against its category and not all records from a table. For example I don’t get any joy in saying I never won the state championship in Greco Roman wrestling. On the other hand with great joy I tell people was the Tacoma City Champion. By partitioning the sport down into City instead of the entire country or state my ranking is much better.

Businesses might have other reasons for partitioning a group. Let’s use an example of schools where they see most of them getting a drop in attendance during the summer time. Comparing an attendance trend from spring to summer would not show an uptrend even if your school is growing year after year. It makes more sense to compare this summer to last summer to find a trend then to compare the spring and summer where you always see a drop. Whatever your reason you can get a total by a level of your choosing with the OVER() clause. Now let’s try an example using the aggregate function COUNT( ) with the OVER( ) clause. We’ll start with a simple query on the Employee table. Here we see Alex Adams is just 1 employee out of 13.  In fact each employee is one of 13 in this example.

SQL SERVER - Tips from the SQL Joes 2 Pros Development Series - Advanced Aggregates with the Over Clause - Day 11 of 35 j2p_11_1

Now let’s look at Alex Adams. He works in Location 1. How many of JProCo’s employees are in Location 1 (Seattle)? And how many of the total employees work in Boston? Alongside the existing table records, we want to add a column showing the count of employees at each location. Recall that blank parentheses cause OVER( ) to apply the aggregation across all rows of a query. Our instruction to show employee count at each location means we don’t just want a total of all locations in our query – we already have that data displayed in the TotalEmployees column. We must include an argument inside the parentheses. PARTITION BY divides the result set into partitions. In this case we’ve added a partition by LocationID.

SQL SERVER - Tips from the SQL Joes 2 Pros Development Series - Advanced Aggregates with the Over Clause - Day 11 of 35 j2p_11_2

Similar to the behavior of GROUP BY, the command to PARTITION BY LocationID takes the LocationID for each employee and counts the number of records in that group which shares the same LocationID. Thus, each of the Seattle employees shows a 7 for LocationCount. The Boston employees have a LocationCount of 3, and the Spokane records have LocationCount of 2. John Marshbank shows LocationCount of 1, since he’s the only one with a null LocationID.

Since Seattle has 7 of the total 13 employees, we can guesstimate that just over ½ (or more than 50%) of JProCo employees work in Seattle (LocationID 1). Let’s run the query and check the results.

SQL SERVER - Tips from the SQL Joes 2 Pros Development Series - Advanced Aggregates with the Over Clause - Day 11 of 35 j2p_11_1

Our result set shows no ratios and all 0’s. This is because our calculation involved only integers. (The grant amount data was expressed in decimal form.) Rather than writing additional code using CAST, we’ll use a trick that works with many programming languages. Multiply the numerator by 1.0 to make each figure display as a decimal instead of an integer. Multiply the numerator by 1.0, so results appear as decimals instead of integers.

SQL SERVER - Tips from the SQL Joes 2 Pros Development Series - Advanced Aggregates with the Over Clause - Day 11 of 35 j2p_11_4

This result is a big improvement – each result is now in decimal form. The Seattle occupancy rate is about 53.85%.  Now multiply instead by 100 so the values show as percentages, like we did with the Grant example. In this instance be sure to multiply by 100.0, so the values show as decimals and not integers. 

Finally, polish your report by adding the descriptive title Pct in place of the blank column header. Sort your report in descending order of occupancy (Pct) and LastName.

SQL SERVER - Tips from the SQL Joes 2 Pros Development Series - Advanced Aggregates with the Over Clause - Day 11 of 35 j2p_11_5

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 SQLQueriesChapter5.0Setup.sql script from Volume 2.

Question 11

Q 11) You have a table named CurrentProducts. The table contains a column named Category. You need to make a T-SQL statement that calculates the percentage (with decimals) of products in each product Category. Which query should you use?

  1. SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts
  2. SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts
  3. SELECT DISTINCT Category,
    COUNT(*) OVER ( )*100.0/ COUNT(*) OVER(PARTITION BY Category) as PctCategory
    FROM CurrentProducts

Rules:

Please leave your answer in comment section below with correct option, explanation and your country of resident.
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A valid answer must contain country of residence of answerer.
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Reference:  Pinal Dave (https://blog.sqlauthority.com)

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108 Comments. Leave new

  • Correct answer is option #2.

    2.SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    – as expained clearly in the aritcle above.

    Option #1 displays categories with percentages in integer values.
    Option #2 displays categories with percentages in decimal values.
    Option #3 displays incorrect results.

    Thanks.

    Country – India

    Reply
  • Rene Alberto Castro Velasquez
    August 11, 2011 7:45 am

    Correct answer is No. 2, because the OVER() includes a PARTITION BY clause and uses a multiplication by 100.0 to show results as percentage.
    Rene Castro
    El Salvador

    Reply
  • Shatrughna Kumar
    August 11, 2011 7:46 am

    Correct option is 2.
    New Delhi

    Reply
  • Hi

    The answer is option 2
    Reason we are counting for each category * 100.0 for making the result to float and / count * to divide by total count we will get category wise percentage for each category
    Sathya – India

    Reply
  • Option 2 is correct because we need to partition by Category and have result in decimal places so 100.0 must be used.

    Thanks

    Sudhir Chawla
    New delhi, India

    Reply
  • Basavaraj Biradar
    August 11, 2011 8:26 am

    Correct Answer is 2

    Thanks,
    Basavaraj

    Reply
  • Aditya Bisoi (@AdityaBisoi07)
    August 11, 2011 8:55 am

    Question 11

    Ans : SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Chennai, INDIA

    Reply
  • Uday Kumar B R
    August 11, 2011 9:09 am

    Question 11

    Q 11) You have a table named CurrentProducts. The table contains a column named Category. You need to make a T-SQL statement that calculates the percentage (with decimals) of products in each product Category. Which query should you use?

    Answer is :2 Gives the result in decimal with correct results
    2. SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Wrong :As we expecting resulst in decimal, following query wont give results in decimal
    1. SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Wrong:Before dividing we are using COUNT(*) OVER ( )*100.0–>this will takes the total count of all rows as over clause is not partitioned and
    COUNT(*) OVER(PARTITION BY Category) —>divider will be less as we have specified the PARTITION BY Category for count ..This will give invalid result.
    3. SELECT DISTINCT Category,
    COUNT(*) OVER ( )*100.0/ COUNT(*) OVER(PARTITION BY Category) as PctCategory
    FROM CurrentProducts

    Thanks for the post :-)

    Country: India

    Reply
  • Option 2:
    SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Option 1 has results in integer without the decimals

    Option 3 is wrong as it has total product count in the numerator

    Leo PIus
    USA

    Reply
  • Correct answer is option 2

    SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Sumit

    India

    Reply
  • Partha Pratim Dinda
    August 11, 2011 10:13 am

    Ans is 2 option
    2.SELECT DISTINCT Category,
    COUNT(*) OVER ( )*100.0/ COUNT(*) OVER(PARTITION BY Category) as PctCategory
    FROM CurrentProducts

    Partha
    India

    Reply
  • Option 2 is correct.

    2.SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    As Option 1 is nearly correct but it will return value in integer.
    Option 3 is incorrect any how.

    INDIA

    Reply
  • Correct answer is option 2 that is

    SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Because we need to display percentage (with decimals) of products in each product Category we need first to find two things
    1. No of product in particular category
    2. Total no of products in the table
    and also we need result in decimal point we have to use COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() .

    Mahmad Khoja
    INDIA
    AHMEDABAD

    Reply
  • Nagaraj Ejanthkar
    August 11, 2011 10:29 am

    The answer to this question is 2nd option.

    SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Option 1: Gives the Percentage of products since it is partitioned by Category without the decimals over the total count.
    Option 2: Gives the Percentage of products since it is partitioned by Category with the decimals over the total count.
    Option 3:This query doesn’t give the percentage but gives the value of total count over count per category.

    Nagaraj Ejanthkar
    USA

    Reply
  • Varinder Sandhu
    August 11, 2011 10:38 am

    Option 2 is correct answer as

    SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    because it displays categories with percentages in decimal values.

    Varinder Sandhu (India)

    Reply
  • Answer No. 2 is correct

    The OVER() includes a PARTITION BY clause with Category which count the records category wise and uses a multiplication by 100.0 and division by all records of the table to show results as percentage.

    India

    Reply
  • Answer is :

    SELECT DISTINCT Category,
    COUNT(*) OVER ( )*100.0/ COUNT(*) OVER(PARTITION BY Category) as PctCategory
    FROM CurrentProducts

    Reply
  • Answer is 2:

    SELECT DISTINCT Category,
    COUNT(*) OVER ( )*100.0/ COUNT(*) OVER(PARTITION BY Category) as PctCategory
    FROM CurrentProducts

    because: Answer 1 will give integer value,
    answer 2 will give exact decimal points…

    Chennai
    India

    Reply
  • Answer is 2:

    SELECT DISTINCT Category,
    COUNT(*) OVER ( )*100.0/ COUNT(*) OVER(PARTITION BY Category) as PctCategory
    FROM CurrentProducts

    because: Answer 1 is giving integer,
    answer 2 will give exact decimal points…

    Chennai
    India

    Reply
  • kalyanasundaram.k
    August 11, 2011 11:01 am

    Answer for today blog quiz : Option 2

    2) SELECT DISTINCT Category,
    COUNT(*) OVER (PARTITION BY Category)*100.0/ COUNT(*) OVER() as PctCategory
    FROM CurrentProducts

    Chennai, TamilNadu, India

    Reply

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