Table: Accounts
+-------------+------+ | Column Name | Type | +-------------+------+ | account_id | int | | income | int | +-------------+------+ account_id is the primary key (column with unique values) for this table. Each row contains information about the monthly income for one bank account.
Write a solution to calculate the number of bank accounts for each salary category. The salary categories are:
"Low Salary": All the salaries strictly less than$20000."Average Salary": All the salaries in the inclusive range[$20000, $50000]."High Salary": All the salaries strictly greater than$50000.
The result table must contain all three categories. If there are no accounts in a category, return 0.
Return the result table in any order.
The result format is in the following example.
Example 1:
Input: Accounts table: +------------+--------+ | account_id | income | +------------+--------+ | 3 | 108939 | | 2 | 12747 | | 8 | 87709 | | 6 | 91796 | +------------+--------+ Output: +----------------+----------------+ | category | accounts_count | +----------------+----------------+ | Low Salary | 1 | | Average Salary | 0 | | High Salary | 3 | +----------------+----------------+ Explanation: Low Salary: Account 2. Average Salary: No accounts. High Salary: Accounts 3, 6, and 8.
# Write your MySQL query statement below
WITH
S AS (
SELECT 'Low Salary' AS category
UNION
SELECT 'Average Salary'
UNION
SELECT 'High Salary'
),
T AS (
SELECT
CASE
WHEN income < 20000 THEN "Low Salary"
WHEN income > 50000 THEN 'High Salary'
ELSE 'Average Salary'
END AS category,
COUNT(1) AS accounts_count
FROM Accounts
GROUP BY category
)
SELECT s.category, IFNULL(accounts_count, 0) AS accounts_count
FROM
S AS s
LEFT JOIN T AS t ON s.category = t.category;