TranslateProject/sources/tech/20220902 How to display the presence and absence of nth-highest group-wise values in SQL.md

[#]: subject: "How to display the presence and absence of nth-highest group-wise values in SQL"
[#]: via: "https://opensource.com/article/22/9/nth-highest-values-sql"
[#]: author: "Mohammed Kamil Khan https://opensource.com/users/kamilk98"
[#]: collector: "lkxed"
[#]: translator: " "
[#]: reviewer: " "
[#]: publisher: " "
[#]: url: " "

How to display the presence and absence of nth-highest group-wise values in SQL
======
A step-by-step breakdown of the query.

![Digital creative of a browser on the internet][1]

While skimming through SQL to prepare for interviews, I often come across this question: Find the employee with the highest or (second-highest) salary by joining a table containing employee information with another that contains department information. This raises a further question: What about finding the employee who earns the nth-highest salary department-wide?

Now I want to pose a more complex scenario: What will happen when a department doesn't have an employee earning the nth-highest salary? For example, a department with only two employees will not have an employee earning the third-highest salary.

Here's my approach to this question:

### Create department and employee tables

I create a table that includes fields such as `dept_id` and `dept_name`.

```
CREATE TABLE department (
    dept_id INT,
    dept_name VARCHAR(60)
);
```

Now I insert various departments into the new table.

```
INSERT INTO department (dept_id,dept_name)
VALUES (780,'HR');
INSERT INTO department (dept_id,dept_name)
VALUES (781,'Marketing');
INSERT INTO department (dept_id,dept_name)
VALUES (782,'Sales');
INSERT INTO department (dept_id,dept_name)
VALUES (783,'Web Dev');
```

![A table showing the data from the earlier code snippets with the columns "Department ID" and "Department Name"][2]

igure 1. The department table 

Next, I create another table incorporating the fields `first_name`, `last_name`, `dept_id`, and `salary`.

```
CREATE TABLE employee (
    first_name VARCHAR(100),
    last_name VARCHAR(100),
    dept_id INT,
    salary INT
);
```

Then I insert values into the table:

```
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Sam','Burton',781,80000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Peter','Mellark',780,90000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Happy','Hogan',782,110000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Steve','Palmer',782,120000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Christopher','Walker',783,140000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Richard','Freeman',781,85000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Alex','Wilson',782,115000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Harry','Simmons',781,90000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Thomas','Henderson',780,95000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Ronald','Thompson',783,130000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('James','Martin',783,135000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Laurent','Fisher',780,100000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Tom','Brooks',780,85000);
INSERT INTO employee (first_name,last_name,dept_id,salary)
VALUES ('Tom','Bennington',783,140000);
```

![A table showing data from the earlier code snippets with first name, last name, dept ID, and salary columns, ordered by department ID number][3]

Figure 2. A table of employees ordered by department ID 

I can infer the number of employees in each department using this table (department ID:number of employees):

* 780:4
* 781:3
* 782:3
* 783:4

If I want the view the second-highest-earning employees from different departments, along with their department's name (using `DENSE_RANK` ), the table will be as follows:

![A table with department ID, department name, first name, last name, and salary columns, listing the second-highest-earning employee in each of four departments, ordered from lowest to highest salary][4]

Figure 3. The second-highest-earning employee in each department 

If I apply the same query to find the fourth-highest-earning employees, the output will be only Tom Brooks of department 780 (HR), with a salary of $85,000.

![The table listing fourth-highest-earning employees lists only one employee.][5]

Figure 4. The fourth-highest-earning employee 

Though department 783 (Web Dev) has four employees, two (James Martin and Ronald Thompson) will be classified as the third-highest-earning employees of that department, since the top two earners have the same salary.

### Finding the nth highest

Now, to the main question: What if I want to display the `dept_ID` and `dept_name` with null values for employee-related fields for departments that do not have an nth-highest-earning employee?

![The list of fourth-highest-earning employee by department, showing "null" in the first name, last name, and salary columns for departments that do not have a fourth-highest earner.][6]

Figure 5. All departments listed, whether or not they have an nth-highest-earning employee 

The table displayed in Figure 5 is what I am aiming to obtain when specific departments do not have an nth-highest-earning employee: The marketing, sales, and web dev departments are listed, but the name and salary fields contain a null value.

The ultimate query that helps obtain the table in Figure 5 is as follows:

```
SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
FROM (SELECT * FROM (
SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
FROM employee INNER JOIN department
ON employee.dept_id=department.dept_id) AS k
WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT NULL)),
nulled AS(SELECT
CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
first_name,last_name,salary
FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
SELECT * FROM null1
UNION
SELECT * FROM nulled
ORDER BY dept_id)
B;
```

### Breakdown of the query

I will break down the query to make it less overwhelming.

Use `DENSE_RANK()` to display employee and department information (not involving null for the absence of the nth-highest-earning member):

```
SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4
```

Output:

![A table of the fourth-highest earners showing only the department with a fourth-highest earner][7]

Figure 6. The fourth-highest earner 

Exclude the `rank1` column from the table in Figure 6, which identifies only one employee with a fourth-highest salary, even though there are four employees in another department.

```
SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    FROM (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A
```

Output:

![The fourth-highest earner table (table six) without the rank 1 column][8]

Figure 7. The fourth-highest earner table without the rank 1 column 

Point out the departments from the department table that do not have an nth-highest-earning employee:

```
SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    FROM (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT NULL))
SELECT * FROM full1)B
```

Output:

![The full1 table listing the departments without a fourth-highest earner by department ID and name: marketing, sales, web dev][9]

Figure 8. The full1 table listing the departments without a fourth-highest earner 

Replace `full1` in the last line of the above code with `null1` :

```
SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    FROM (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT NULL))
SELECT * FROM null1)B
```

![The null1 table listing all departments, with null values for those without a fourth-highest earner][10]

Figure 9. The null1 table listing all departments, with null values for those without a fourth-highest earner 

Now, I fill the null values of the `dept_id` and `dept_name` fields in Figure 9 with the corresponding values from Figure 8.

```
SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
    FROM (SELECT * FROM (
  SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
   employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
   FROM employee INNER JOIN department
   ON employee.dept_id=department.dept_id) AS k
   WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT NULL)),
nulled AS(SELECT
CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
first_name,last_name,salary
FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
SELECT * FROM nulled) B;
```

![The table with department id, department name, first name, last name, and salary columns, with null values in the name and salary columns][11]

Figure 10. The result of the nulled query 

The nulled query uses `CASE WHEN` on the nulls encountered in the `dept_id` and `dept_name` columns of the `null1` table and replaces them with the corresponding values in the `full1` table. Now all I need to do is apply `UNION` to the tables obtained in Figure 7 and Figure 10. This can be accomplished by declaring the last query in the previous code using `WITH` and then `UNION` -izing it with `null1`.

```
SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
FROM (SELECT * FROM (
SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
FROM employee INNER JOIN department
ON employee.dept_id=department.dept_id) AS k
WHERE rank1=4)A),
full1 AS (SELECT dept_id, dept_name FROM department WHERE dept_id NOT IN (SELECT dept_id FROM null1 WHERE dept_id IS NOT NULL)),
nulled AS(SELECT
CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
first_name,last_name,salary
FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
SELECT * FROM null1
UNION
SELECT * FROM nulled
ORDER BY dept_id)
B;
```

![The complete table: department ID, department name, first name, last name, salary columns. The first row contains the information of the one fourth-highest earner, and the next three columns show the remaining departments, with ID, and null value in the other three columns.][12]

Figure 11. The final result 

Now I can infer from Figure 11 that marketing, sales, and web dev are the departments that do not have any employees earning the fourth-highest salary.

Image By: (Mohammed Kamil Khan, CC BY-SA 4.0)

--------------------------------------------------------------------------------

via: https://opensource.com/article/22/9/nth-highest-values-sql

作者：[Mohammed Kamil Khan][a]
选题：[lkxed][b]
译者：[译者ID](https://github.com/译者ID)
校对：[校对者ID](https://github.com/校对者ID)

本文由 [LCTT](https://github.com/LCTT/TranslateProject) 原创编译，[Linux中国](https://linux.cn/) 荣誉推出

[a]: https://opensource.com/users/kamilk98
[b]: https://github.com/lkxed
[1]: https://opensource.com/sites/default/files/lead-images/browser_web_internet_website.png
[2]: https://opensource.com/sites/default/files/2022-08/fig.%201%20sql.png
[3]: https://opensource.com/sites/default/files/2022-08/FIG%202%20sql.png
[4]: https://opensource.com/sites/default/files/2022-08/fig%203%20sql.png
[5]: https://opensource.com/sites/default/files/2022-08/fg%204%20sql.png
[6]: https://opensource.com/sites/default/files/2022-08/fig%205%20sql.png
[7]: https://opensource.com/sites/default/files/2022-08/fig%206%20sql.png
[8]: https://opensource.com/sites/default/files/2022-08/fig%207%20sql.png
[9]: https://opensource.com/sites/default/files/2022-08/fig%208%20sql.png
[10]: https://opensource.com/sites/default/files/2022-08/fig%209%20sql.png
[11]: https://opensource.com/sites/default/files/2022-08/fig%2010%20sql.png
[12]: https://opensource.com/sites/default/files/2022-08/fig%2011%20sql.png