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SQL Challenges-1: Sales Analysis

SQL Challenges-1: Exercise-35 with Solution

From the following tables write an SQL query to find the best seller by total sales price. Return distributor ID , If there is a tie, report them all.

Input:

Table: item

Structure:

FieldTypeNullKeyDefaultExtra
item_codeint(11)NOPRI
item_descvarchar(255)YES
costint(11)YES

Data:

item_codeitem_desccost
101mother board2700
102RAM800
103key board300
104mouse300

Table: sales_info

Structure:

FieldTypeNullKeyDefaultExtra
distributor_idint(11)YES
item_codeint(11)YES
retailer_id int(11)YES
date_of_selldateYES
quantityint(11)YES
total_costint(11)YES

Data:

distributor_iditem_coderetailer_iddate_of_sellquantitytotal_cost
500110110012020-02-1238100
500110310022020-03-15154500
500210110012019-06-2425400
500110410032019-09-1182400
500310110032020-10-21513500
500310410022020-12-27103000
500210210012019-05-18129600
500210310042020-06-1782400
500310310012020-04-123900

Sample Solution-1:

SQL Code(MySQL):

CREATE TABLE item (item_code int not null unique, item_desc varchar(255), cost int);
INSERT INTO item VALUES(101,'mother board',	2700);
INSERT INTO item VALUES(102,'RAM',	800);
INSERT INTO item VALUES(103,'key board',300);
INSERT INTO item VALUES(104,'mouse',300);


CREATE TABLE sales_info (distributor_id int, item_code int, retailer_id int, date_of_sell date, quantity int, total_cost int);
INSERT INTO sales_info VALUES(5001,101,1001,'2020-02-12',3,8100);
INSERT INTO sales_info VALUES(5001,103,1002,'2020-03-15',15,4500);
INSERT INTO sales_info VALUES(5002,101,1001,'2019-06-24',2,5400);
INSERT INTO sales_info VALUES(5001,104,1003,'2019-09-11',8,2400);
INSERT INTO sales_info VALUES(5003,101,1003,'2020-10-21',5,13500);
INSERT INTO sales_info VALUES(5003,104,1002,'2020-12-27',10,3000);
INSERT INTO sales_info VALUES(5002,102,1001,'2019-05-18',12,9600);
INSERT INTO sales_info VALUES(5002,103,1004,'2020-06-17',8,2400);
INSERT INTO sales_info VALUES(5003,103,1001,'2020-04-12',3,900);

SELECT DISTINCT distributor_id
FROM sales_info
GROUP BY distributor_id
HAVING SUM(total_cost) =
(
SELECT SUM(total_cost) AS total_price
FROM sales_info
GROUP BY distributor_id
ORDER BY total_price DESC
LIMIT 1
);

Sample Output:

distributor_id|
--------------|
          5002|
          5003|

OR

Sample Solution-2:

SQL Code(MySQL):

CREATE TABLE item (item_code int not null unique, item_desc varchar(255), cost int);
INSERT INTO item VALUES(101,'mother board',	2700);
INSERT INTO item VALUES(102,'RAM',	800);
INSERT INTO item VALUES(103,'key board',300);
INSERT INTO item VALUES(104,'mouse',300);


CREATE TABLE sales_info (distributor_id int, item_code int, retailer_id int, date_of_sell date, quantity int, total_cost int);
INSERT INTO sales_info VALUES(5001,101,1001,'2020-02-12',3,8100);
INSERT INTO sales_info VALUES(5001,103,1002,'2020-03-15',15,4500);
INSERT INTO sales_info VALUES(5002,101,1001,'2019-06-24',2,5400);
INSERT INTO sales_info VALUES(5001,104,1003,'2019-09-11',8,2400);
INSERT INTO sales_info VALUES(5003,101,1003,'2020-10-21',5,13500);
INSERT INTO sales_info VALUES(5003,104,1002,'2020-12-27',10,3000);
INSERT INTO sales_info VALUES(5002,102,1001,'2019-05-18',12,9600);
INSERT INTO sales_info VALUES(5002,103,1004,'2020-06-17',8,2400);
INSERT INTO sales_info VALUES(5003,103,1001,'2020-04-12',3,900);

SELECT distributor_id
FROM sales_info
GROUP BY distributor_id
HAVING SUM(total_cost) >= 
ALL(SELECT SUM(total_cost) FROM sales_info GROUP BY distributor_id);

SQL Code Editor:


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Next: Retailers buying history.



SQL: Tips of the Day

How to select the nth row in a SQL database table?

Basically, PostgreSQL and MySQL supports the non-standard:

SELECT...
LIMIT y OFFSET x 

Oracle, DB2 and MSSQL supports the standard windowing functions:

SELECT * FROM (
  SELECT
    ROW_NUMBER() OVER (ORDER BY key ASC) AS rownumber,
    columns
  FROM tablename
) AS foo
WHERE rownumber <= n

Database: SQL

Ref: https://bit.ly/3zPxcD8