SQL Challenges-1: Department wise number of male and female employees and their total salaries

Last update on August 19 2022 21:50:37 (UTC/GMT +8 hours)

SQL Challenges-1: Exercise-56 with Solution

From the following table write a SQL query to find the number of male and female employees in each department and along with their total salaries. Return department ID, number of female employees, their total salaries, number of male employees and their total salaries.

Input:

Table: employees

Structure:

Field	Type	Null	Key
emp_id	int	NO	PRI
emp_name	varchar(30)	YES
emp_sex	varchar(1)	YES
emp_salary	int	YES
emp_department	int	YES

Data:

emp_id	emp_name	emp_sex	emp_salary	emp_department
100	Steven	M	24000	90
101	Neena	F	17000	90
102	Lex	M	17000	80
103	Alexander	M	9000	60
104	Bruce	M	6000	60
105	David	M	4800	80
106	Valli	F	4800	60
107	Diana	F	4200	60
108	Nancy	M	12000	100
109	Daniel	F	9000	100
110	John	M	8200	100
111	Ismael	M	7700	100
112	Jose Manuel	M	7800	100
113	Luis	F	6900	100
114	Den	M	11000	30
115	Alexander	M	3100	30
116	Shelli	F	2900	30
117	Sigal	F	2800	30
133	Jason	M	3300	50
134	Michael	F	2900	50
135	Ki	F	2400	50

Sample Solution:

SQL Code(MySQL):

create table employees (
emp_id integer(4) not null unique,
emp_name varchar(30),
emp_sex varchar(1),
emp_salary int(6),
emp_department int(3));



insert into employees values(100,'Steven     ','M',24000,  90);
insert into employees values(101,'Neena      ','F',17000,  90);
insert into employees values(102,'Lex        ','M',17000,  80);
insert into employees values(103,'Alexander  ','M', 9000,  60);
insert into employees values(104,'Bruce      ','M', 6000,  60);
insert into employees values(105,'David      ','M', 4800,  80);
insert into employees values(106,'Valli      ','F', 4800,  60);
insert into employees values(107,'Diana      ','F', 4200,  60);
insert into employees values(114,'Den        ','M',11000,  30);
insert into employees values(115,'Alexander  ','M', 3100,  30);
insert into employees values(116,'Shelli     ','F', 2900,  30);
insert into employees values(117,'Sigal      ','F', 2800,  30);
insert into employees values(108,'Nancy      ','M',12000, 100);
insert into employees values(109,'Daniel     ','F', 9000, 100);
insert into employees values(110,'John       ','M', 8200, 100);
insert into employees values(111,'Ismael     ','M', 7700, 100);
insert into employees values(112,'Jose Manuel','M', 7800, 100);
insert into employees values(113,'Luis       ','F', 6900, 100);
insert into employees values(133,'Jason      ','M', 3300,  50);
insert into employees values(134,'Michael    ','F', 2900,  50);
insert into employees values(135,'Ki         ','F', 2400,  50);



SELECT  emp_department AS department,
SUM(CASE WHEN emp_sex='F' THEN 1 END) AS female_employees,
SUM(CASE WHEN emp_sex='F' THEN emp_salary END) AS female_total_salary,
SUM(CASE WHEN emp_sex='M' THEN 1 END) AS male_employees,
SUM(CASE WHEN emp_sex='M' THEN emp_salary END) AS male_total_salary
FROM employees
group BY 1;

Sample Output:

department|female_employees|female_total_salary|male_employees|male_total_salary|
----------+----------------+-------------------+--------------+-----------------+
        90|               1|              17000|             1|            24000|
        80|                |                   |             2|            21800|
        60|               2|               9000|             2|            15000|
       100|               2|              15900|             4|            35700|
        30|               2|               5700|             2|            14100|
        50|               2|               5300|             1|             3300|

SQL Code Editor:

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