Mysql 如何在sql中包含0作为计数

表:帐目

+-------------+------+
| Column Name | Type |
+-------------+------+
| account_id  | int  |
| income      | int  |
+-------------+------+

Account_id是该表的主键(具有唯一值的列). 每一行都包含有关一个银行帐户的月收入的信息.

写一个解决方案来计算每个薪资类别的银行账户数量.薪金类别包括:

• "低工资":所有工资严格低于20000美元."平均水平
• 薪金":包括范围内的所有薪金[20000美元、50000美元].
• "高薪":所有薪水严格高于50000美元.

结果表必须包含所有三个类别.如果没有 类别中的帐户，则返回0.

以任意顺序返回结果表.

结果格式如下例所示.

例1:

输入: 帐户表:

+------------+--------+
| account_id | income |
+------------+--------+
| 3          | 108939 |
| 2          | 12747  |
| 8          | 87709  |
| 6          | 91796  |
+------------+--------+

输出:

+----------------+----------------+
| category       | accounts_count |
+----------------+----------------+
| Low Salary     | 1              |
| Average Salary | 0              |
| High Salary    | 3              |
+----------------+----------------+

解释: 低薪:账号2. 平均工资:没有账户. 高薪:账户3、6和8.

link to the problem statement. I tried this.

SELECT 
CASE 
WHEN income < 20000 THEN "Low Salary"
WHEN income >= 20000 AND income <= 50000 THEN "Average Salary"
ELSE "High Salary"
END AS category,
COUNT(*) AS accounts_count
FROM Accounts 
GROUP BY category;

My output is this个

| category    | accounts_count |
| ----------- | -------------- |
| High Salary | 3              |
| Low Salary  | 1              |

But expected output is this个

| category       | accounts_count |
| -------------- | -------------- |
| High Salary    | 3              |
| Low Salary     | 1              |
| Average Salary | 0              |