SQL Group By Number Of Users Within Range - mysql

Question

I have the following query which will return the number of users in table transactions who have earned between $100 and $200
SELECT COUNT(users.id)
FROM transactions
LEFT JOIN users ON users.id = transactions.user_id
WHERE transactions.amount > 100 AND transactions.amount < 200
The above query returns the correct result below:
COUNT(users.id)
559
I would like to extend it so that the query can return data in the following format:
COUNT(users.id) : amount
1678 : 0-100
559 : 100-200
13 : 200-300
How can I do this?

Answer 1

You can use a CASE expression inside of your aggregate function which will get the result in columns:
SELECT
COUNT(case when amount >= 0 and amount <= 100 then users.id end) Amt0_100,
COUNT(case when amount >= 101 and amount <= 200 then users.id end) Amt101_200,
COUNT(case when amount >= 201 and amount <= 300 then users.id end) Amt201_300
FROM transactions
LEFT JOIN users
ON users.id = transactions.user_id;
See SQL Fiddle with Demo
You will notice that I altered the ranges from 0-100, 101-200, 201-300 otherwise you will have user ids being counted twice on the 100, 200 values.
If you want the values in rows, then you can use:
select count(u.id),
CASE
WHEN amount >=0 and amount <=100 THEN '0-100'
WHEN amount >=101 and amount <=200 THEN '101-200'
WHEN amount >=201 and amount <=300 THEN '101-300'
END Amount
from transactions t
left join users u
on u.id = t.user_id
group by
CASE
WHEN amount >=0 and amount <=100 THEN '0-100'
WHEN amount >=101 and amount <=200 THEN '101-200'
WHEN amount >=201 and amount <=300 THEN '101-300'
END
See SQL Fiddle with Demo
But if you have many ranges that you need to calculate the counts on, then you might want to consider creating a table with the ranges, similar to the following:
create table report_range
(
start_range int,
end_range int
);
insert into report_range values
(0, 100),
(101, 200),
(201, 300);
Then you can use this table to join to your current tables and group by the range values:
select count(u.id) Total, concat(start_range, '-', end_range) amount
from transactions t
left join users u
on u.id = t.user_id
left join report_range r
on t.amount >= r.start_range
and t.amount<= r.end_range
group by concat(start_range, '-', end_range);
See SQL Fiddle with Demo.
If you don't want to create a new table with the ranges, then you can always use a derived table to get the same result:
select count(u.id) Total, concat(start_range, '-', end_range) amount
from transactions t
left join users u
on u.id = t.user_id
left join
(
select 0 start_range, 100 end_range union all
select 101 start_range, 200 end_range union all
select 201 start_range, 300 end_range
) r
on t.amount >= r.start_range
and t.amount<= r.end_range
group by concat(start_range, '-', end_range);
See SQL Fiddle with Demo

Answer 2

One way to do this would be to use a case/when statement in your group by.
SELECT
-- NB this must match your group by statement exactly
-- otherwise you will get an error
CASE
WHEN amount <= 100
THEN '0-100'
WHEN amount <= 200
THEN '100-200'
ELSE '201+'
END Amount,
COUNT(*)
FROM
transactions
GROUP BY
CASE
WHEN amount <= 100
THEN '0-100'
WHEN amount <= 200
THEN '100-200'
ELSE '201+'
END
If you plan on using the grouping elsewhere, it probably makes sense to define it as a scalar function (it will also look cleaner)
e.g.
SELECT
AmountGrouping(amount),
COUNT(*)
FROM
transactions
GROUP BY
AmountGrouping(amount)
If you want to be fully generic:
SELECT
concat(((amount DIV 100) * 100),'-',(((amount DIV 100) + 1) * 100)) AmountGroup,
COUNT(*)
FROM
transactions
GROUP BY
AmountGroup
Sql Fiddle

Answer 3

Bilbo, I tried to be creative and found a very nice solution [ for those who love math (like me) ]
It's always surprising when MySQL integer division operator solves our problems.
DROP SCHEMA IF EXISTS `stackoverflow3`;
CREATE SCHEMA `stackoverflow3`;
USE `stackoverflow3`;
CREATE TABLE users (
id INT UNSIGNED PRIMARY KEY NOT NULL AUTO_INCREMENT,
name VARCHAR(25) NOT NULL DEFAULT "-");
CREATE TABLE transactions(
id INT UNSIGNED PRIMARY KEY NOT NULL AUTO_INCREMENT,
user_id INT UNSIGNED NOT NULL,
amount INT UNSIGNED DEFAULT 0,
FOREIGN KEY (user_id) REFERENCES users (id));
INSERT users () VALUES (),(),();
INSERT transactions (user_id,amount)
VALUES (1,120),(2,270),(3, 350),
(2,500), (1,599), (1,550), (3,10),
(3,20), (3,30), (3,50), (3,750);
SELECT
COUNT(t.id),
CONCAT(
((t.amount DIV 100)*100)," to ",((t.amount DIV 100 + 1)*100-1)
) AS amount_range
FROM transactions AS t
GROUP BY amount_range;
Awaiting your questions, Mr. Baggins.