From c082f8abf9cec27a343fee4725f963d155bfee05 Mon Sep 17 00:00:00 2001 From: lkxed Date: Fri, 2 Sep 2022 21:34:08 +0800 Subject: [PATCH] =?UTF-8?q?[=E6=89=8B=E5=8A=A8=E9=80=89=E9=A2=98][tech]:?= =?UTF-8?q?=2020220902=20How=20to=20display=20the=20presence=20and=20absen?= =?UTF-8?q?ce=20of=20nth-highest=20group-wise=20values=20in=20SQL.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...of nth-highest group-wise values in SQL.md | 307 ++++++++++++++++++ 1 file changed, 307 insertions(+) create mode 100644 sources/tech/20220902 How to display the presence and absence of nth-highest group-wise values in SQL.md diff --git a/sources/tech/20220902 How to display the presence and absence of nth-highest group-wise values in SQL.md b/sources/tech/20220902 How to display the presence and absence of nth-highest group-wise values in SQL.md new file mode 100644 index 0000000000..a9e8017b3c --- /dev/null +++ b/sources/tech/20220902 How to display the presence and absence of nth-highest group-wise values in SQL.md @@ -0,0 +1,307 @@ +[#]: 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