CODEWITHSUNDEEP
mysqlsql
rtorres
rtorres

Reputation: 2951

SQL Query to count rows that have a common column value based on another column value

I have a single MySQL table with this data: 

CREATE TABLE job_history(
  id INT PRIMARY KEY AUTO_INCREMENT,
  employee VARCHAR(50),
  company VARCHAR(50)
);

INSERT INTO job_history(employee, company) VALUES
('John', 'IBM'),
('John', 'Walmart'),
('John', 'Uber'),
('Sharon', 'IBM'),
('Sharon', 'Uber'),
('Matt', 'Walmart'),
('Matt', 'Starbucks'),
('Carl', 'Home Depot');

SELECT * FROM job_history;
+----+----------+------------+
| id | employee | company    |
+----+----------+------------+
|  1 | John     | IBM        |
|  2 | John     | Walmart    |
|  3 | John     | Uber       |
|  4 | Sharon   | IBM        |
|  5 | Sharon   | Uber       |
|  6 | Matt     | Walmart    |
|  7 | Matt     | Starbucks  |
|  8 | Carl     | Home Depot |
+----+----------+------------+

Here's the corresponding SQL Fiddle

I want to create a SQL query to count the number of common companies between a given employee and other employees on the table.

For example, if I wanted to target employee 'John', I expect this result:

Sharon: 2
Matt: 1
Carl: 0

because Sharon has 2 common companies with John (IBM and Uber), Matt has 1 common company with John (Walmart), and Carl has 0 companies common with John.

How can I do that?

Upvotes: 1

Views: 1191

Answers (5)

cdaiga
cdaiga

Reputation: 4937

How can I do that?

Do a self join of the data on different employees, and company, then group the result on the employee and count the rows.

 SELECT B.employee, COUNT(A.company) FROM
 (SELECT  *  FROM JOB_HISTORY WHERE employee='John') A RIGHT JOIN
 (SELECT  *  FROM JOB_HISTORY WHERE employee<>'John') B
 ON A.company=B.company
 GROUP BY B.employee
 ORDER BY COUNT(A.company) DESC;

Upvotes: 0

Khan M
Khan M

Reputation: 415

select 
    count(*) as comon_companies, b.employee 
from 
    job_history a
inner join  
    job_history b on a.company = b.company
where 
    a.employee <> b.employee
    and a.employee = 'sharon'
group by 
    b.employee

Use this query to get your desired result. You only need to replace a.employee value in where clause to input name of your target employee

Upvotes: 0

Prabhath Amaradasa
Prabhath Amaradasa

Reputation: 503

-- This will give the count with matching one

SELECT
employee, COUNT(company)
FROM job_history  WHERE company  IN
(
SELECT
company
FROM job_history  WHERE employee ='John' -- parameter
) AND employee <> 'John'  -- parameter
GROUP BY employee

-- Adding count with all employees  

SELECT DISTINCT 
ac.employee,
 COALESCE(mc.JobCount, 0)  AS JobCount
FROM job_history  AS ac LEFT OUTER JOIN
(
SELECT
employee, COUNT(company) AS JobCount
FROM job_history  WHERE company  IN
(
SELECT
company
FROM job_history  WHERE employee ='John'  -- parameter
) AND employee <> 'John'   -- parameter
GROUP BY employee
 ) AS mc ON mc.employee  = ac.employee
 WHERE ac.employee <> 'John'   -- parameter

Upvotes: 0

Gordon Linoff
Gordon Linoff

Reputation: 1270391

First, you need a left join -- because you want all employees even those with no companies in common. Second, group by to get the count:

select jh.employee, count(jh_john.company) as num_in_common
from job_history jh left join
     job_history jh_john
     on jh_john.company = jh.company and
        jh_john.employee = 'John'
where jh.employee <> 'John'
group by jh.employee;

Note: If there could be duplicates in the table, then use count(distinct) rather than count().

Upvotes: 2

Josh Withee
Josh Withee

Reputation: 11356

Try this:

SELECT jh2.employee, count(*)
FROM job_history jh1, job_history jh2
WHERE jh1.company = jh2.company
AND jh1.employee = 'John'
AND jh2.employee <> 'John'
GROUP BY jh2.employee

And if you want them in order from most companies in common to least, add this to the end of the query:

ORDER BY count(*) DESC

Upvotes: 0

Related Questions