SQL Challenges-1: Most experienced manager to execute the schemes

Last update on October 15 2025 12:47:23 (UTC/GMT +8 hours)

34. Most experienced manager to execute the schemes

From the following tables write a SQL query to find those experienced manager who execute the schemes. Return scheme code and scheme manager ID.

Input:

Table: managing_body

Structure:

Data:

Table: scheme

Structure:

Data:

Sample Solution:

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 scheme_code, p.scheme_manager_id
FROM scheme p JOIN managing_body e
ON p.scheme_manager_id = e.manager_id
WHERE (scheme_code, e.running_years) IN
(SELECT scheme_code, MAX(running_years)
FROM scheme p JOIN managing_body e
ON p.scheme_manager_id = e.manager_id
GROUP BY scheme_code);

Sample Output:

scheme_code|scheme_manager_id|
-----------|-----------------|
       1001|               51|
       1002|               51|
       1003|               51|
       1004|               52|

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