SQL Challenges-1: Average experience for each scheme

Last update on December 20 2024 11:39:27 (UTC/GMT +8 hours)

From the following tables write a SQL query to display those managers who have average experience for each scheme.

Input:

Table: managing_body

Structure:

Field	Type	Null	Key
manager_id	int(11)	NO	PRI
manager_name	varchar(255)	YES
running_years	int(11)	YES

Data:

manager_id	manager_name	running_years
51	James	5
52	Cork	3
53	Paul	4
54	Adam	3
55	Hense	4
56	Peter	2

Table: scheme

Structure:

Field	Type	Null	Key	Default	Extra
scheme_code	int(11)	NO	PRI
scheme_manager_id	int(11)	NO	PRI

Data:

scheme_code	scheme_manager_id
1001	51
1001	53
1001	54
1001	56
1002	51
1002	55
1003	51
1004	52

Sample Solution-1:

SQL Code(MySQL):

CREATE TABLE managing_body (manager_id int NOT NULL UNIQUE, manager_name varchar(255), running_years int);

INSERT INTO managing_body VALUES(51,'James',5);
INSERT INTO managing_body VALUES(52,'Cork',3);
INSERT INTO managing_body VALUES(53,'Paul',4);
INSERT INTO managing_body VALUES(54,'Adam',3);
INSERT INTO managing_body VALUES(55,'Hense',4);
INSERT INTO managing_body VALUES(56,'Peter',2);



CREATE TABLE scheme (scheme_code int NOT NULL , scheme_manager_id int NOT NULL, 
PRIMARY KEY(scheme_code,scheme_manager_id));
INSERT INTO scheme VALUES(1001,	51);
INSERT INTO scheme VALUES(1001,	53);
INSERT INTO scheme VALUES(1001,	54);
INSERT INTO scheme VALUES(1001,	56);
INSERT INTO scheme VALUES(1002,	51);
INSERT INTO scheme VALUES(1002,	55);
INSERT INTO scheme VALUES(1003,	51);
INSERT INTO scheme VALUES(1004,	52);


	   
SELECT s.scheme_code  , 
ROUND(SUM(m.running_years) * 1.0/NULLIF(COUNT(DISTINCT m.manager_id), 0) ,2) AS 'Average year of experience'
FROM scheme s JOIN managing_body m 
ON m.manager_id  = s.scheme_manager_id
GROUP BY s.scheme_code;

Sample Output:

scheme_code|Average year of experience|
-----------|--------------------------|
       1001|                      3.50|
       1002|                      4.50|
       1003|                      5.00|
       1004|                      3.00|

Sample Solution-2:

SQL Code(MySQL):

CREATE TABLE managing_body (manager_id int NOT NULL UNIQUE, manager_name varchar(255), running_years int);

INSERT INTO managing_body VALUES(51,'James',5);
INSERT INTO managing_body VALUES(52,'Cork',3);
INSERT INTO managing_body VALUES(53,'Paul',4);
INSERT INTO managing_body VALUES(54,'Adam',3);
INSERT INTO managing_body VALUES(55,'Hense',4);
INSERT INTO managing_body VALUES(56,'Peter',2);



CREATE TABLE scheme (scheme_code int NOT NULL , scheme_manager_id int NOT NULL, 
PRIMARY KEY(scheme_code,scheme_manager_id));
INSERT INTO scheme VALUES(1001,	51);
INSERT INTO scheme VALUES(1001,	53);
INSERT INTO scheme VALUES(1001,	54);
INSERT INTO scheme VALUES(1001,	56);
INSERT INTO scheme VALUES(1002,	51);
INSERT INTO scheme VALUES(1002,	55);
INSERT INTO scheme VALUES(1003,	51);
INSERT INTO scheme VALUES(1004,	52);


	   
SELECT distinct scheme_code,
ROUND(y / n,2) AS 'Average year of experience'
FROM
(SELECT scheme_code, SUM(running_years) AS y, COUNT(*) AS n 
FROM scheme s 
INNER JOIN managing_body m ON s.scheme_manager_id = m.manager_id 
GROUP BY scheme_code) t;

SQL Code Editor:

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