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[手动选题][tech]: 20220902 How to display the presence and absence of nth-highest group-wise values in SQL.md
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[#]: subject: "How to display the presence and absence of nth-highest group-wise values in SQL"
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[#]: via: "https://opensource.com/article/22/9/nth-highest-values-sql"
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[#]: author: "Mohammed Kamil Khan https://opensource.com/users/kamilk98"
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[#]: collector: "lkxed"
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[#]: translator: " "
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[#]: reviewer: " "
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[#]: publisher: " "
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[#]: url: " "
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How to display the presence and absence of nth-highest group-wise values in SQL
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======
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A step-by-step breakdown of the query.
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![Digital creative of a browser on the internet][1]
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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?
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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.
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Here's my approach to this question:
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### Create department and employee tables
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I create a table that includes fields such as `dept_id` and `dept_name`.
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```
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CREATE TABLE department (
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dept_id INT,
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dept_name VARCHAR(60)
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);
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```
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Now I insert various departments into the new table.
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```
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INSERT INTO department (dept_id,dept_name)
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VALUES (780,'HR');
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INSERT INTO department (dept_id,dept_name)
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VALUES (781,'Marketing');
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INSERT INTO department (dept_id,dept_name)
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VALUES (782,'Sales');
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INSERT INTO department (dept_id,dept_name)
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VALUES (783,'Web Dev');
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```
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![A table showing the data from the earlier code snippets with the columns "Department ID" and "Department Name"][2]
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igure 1. The department table
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Next, I create another table incorporating the fields `first_name`, `last_name`, `dept_id`, and `salary`.
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```
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CREATE TABLE employee (
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first_name VARCHAR(100),
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last_name VARCHAR(100),
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dept_id INT,
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salary INT
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);
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```
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Then I insert values into the table:
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```
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Sam','Burton',781,80000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Peter','Mellark',780,90000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Happy','Hogan',782,110000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Steve','Palmer',782,120000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Christopher','Walker',783,140000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Richard','Freeman',781,85000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Alex','Wilson',782,115000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Harry','Simmons',781,90000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Thomas','Henderson',780,95000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Ronald','Thompson',783,130000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('James','Martin',783,135000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Laurent','Fisher',780,100000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Tom','Brooks',780,85000);
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INSERT INTO employee (first_name,last_name,dept_id,salary)
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VALUES ('Tom','Bennington',783,140000);
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```
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![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]
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Figure 2. A table of employees ordered by department ID
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I can infer the number of employees in each department using this table (department ID:number of employees):
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* 780:4
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* 781:3
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* 782:3
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* 783:4
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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:
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![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]
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Figure 3. The second-highest-earning employee in each department
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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.
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![The table listing fourth-highest-earning employees lists only one employee.][5]
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Figure 4. The fourth-highest-earning employee
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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.
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### Finding the nth highest
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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?
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![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]
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Figure 5. All departments listed, whether or not they have an nth-highest-earning employee
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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.
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The ultimate query that helps obtain the table in Figure 5 is as follows:
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```
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SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
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FROM (SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4)A),
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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)),
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nulled AS(SELECT
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CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
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CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
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first_name,last_name,salary
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FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
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SELECT * FROM null1
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UNION
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SELECT * FROM nulled
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ORDER BY dept_id)
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B;
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```
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### Breakdown of the query
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I will break down the query to make it less overwhelming.
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Use `DENSE_RANK()` to display employee and department information (not involving null for the absence of the nth-highest-earning member):
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```
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SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4
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```
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Output:
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![A table of the fourth-highest earners showing only the department with a fourth-highest earner][7]
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Figure 6. The fourth-highest earner
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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.
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```
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SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
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FROM (SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4)A
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```
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Output:
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![The fourth-highest earner table (table six) without the rank 1 column][8]
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Figure 7. The fourth-highest earner table without the rank 1 column
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Point out the departments from the department table that do not have an nth-highest-earning employee:
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```
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SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
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FROM (SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4)A),
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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))
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SELECT * FROM full1)B
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```
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Output:
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![The full1 table listing the departments without a fourth-highest earner by department ID and name: marketing, sales, web dev][9]
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Figure 8. The full1 table listing the departments without a fourth-highest earner
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Replace `full1` in the last line of the above code with `null1` :
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```
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SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
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FROM (SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4)A),
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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))
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SELECT * FROM null1)B
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```
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![The null1 table listing all departments, with null values for those without a fourth-highest earner][10]
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Figure 9. The null1 table listing all departments, with null values for those without a fourth-highest earner
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Now, I fill the null values of the `dept_id` and `dept_name` fields in Figure 9 with the corresponding values from Figure 8.
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```
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SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
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FROM (SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4)A),
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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)),
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nulled AS(SELECT
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CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
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CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
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first_name,last_name,salary
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FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
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SELECT * FROM nulled) B;
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```
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![The table with department id, department name, first name, last name, and salary columns, with null values in the name and salary columns][11]
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Figure 10. The result of the nulled query
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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`.
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```
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SELECT * FROM (WITH null1 AS (SELECT A.dept_id, A.dept_name, A.first_name, A.last_name, A.salary
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FROM (SELECT * FROM (
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SELECT department.dept_id, department.dept_name, employee.first_name, employee.last_name,
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employee.salary, DENSE_RANK() OVER (PARTITION BY employee.dept_id ORDER BY employee.salary DESC) AS Rank1
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FROM employee INNER JOIN department
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ON employee.dept_id=department.dept_id) AS k
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WHERE rank1=4)A),
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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)),
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nulled AS(SELECT
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CASE WHEN null1.dept_id IS NULL THEN full1.dept_id ELSE null1.dept_id END,
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CASE WHEN null1.dept_name IS NULL THEN full1.dept_name ELSE null1.dept_name END,
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first_name,last_name,salary
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FROM null1 RIGHT JOIN full1 ON null1.dept_id=full1.dept_id)
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SELECT * FROM null1
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UNION
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SELECT * FROM nulled
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ORDER BY dept_id)
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B;
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```
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![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]
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Figure 11. The final result
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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.
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Image By: (Mohammed Kamil Khan, CC BY-SA 4.0)
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--------------------------------------------------------------------------------
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via: https://opensource.com/article/22/9/nth-highest-values-sql
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作者：[Mohammed Kamil Khan][a]
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选题：[lkxed][b]
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译者：[译者ID](https://github.com/译者ID)
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校对：[校对者ID](https://github.com/校对者ID)
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本文由 [LCTT](https://github.com/LCTT/TranslateProject) 原创编译，[Linux中国](https://linux.cn/) 荣誉推出
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[a]: https://opensource.com/users/kamilk98
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[b]: https://github.com/lkxed
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[1]: https://opensource.com/sites/default/files/lead-images/browser_web_internet_website.png
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[2]: https://opensource.com/sites/default/files/2022-08/fig.%201%20sql.png
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[3]: https://opensource.com/sites/default/files/2022-08/FIG%202%20sql.png
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[4]: https://opensource.com/sites/default/files/2022-08/fig%203%20sql.png
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[5]: https://opensource.com/sites/default/files/2022-08/fg%204%20sql.png
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[6]: https://opensource.com/sites/default/files/2022-08/fig%205%20sql.png
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[7]: https://opensource.com/sites/default/files/2022-08/fig%206%20sql.png
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[8]: https://opensource.com/sites/default/files/2022-08/fig%207%20sql.png
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[9]: https://opensource.com/sites/default/files/2022-08/fig%208%20sql.png
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[10]: https://opensource.com/sites/default/files/2022-08/fig%209%20sql.png
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[11]: https://opensource.com/sites/default/files/2022-08/fig%2010%20sql.png
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[12]: https://opensource.com/sites/default/files/2022-08/fig%2011%20sql.png
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