如果task_status表中所有相应的状态都设置为“Completed”，则对SQL表进行计数

Question

我正在学习 SQL 解决 Codewars 任务。我不明白如何计算一列，其中所有相应的状态都设置为相同的值。

这是任务：

一家家庭维护公司使用数据库来跟踪需要在各个房屋执行的任务。数据库有两个表：house_tasks 和task_status。 house_tasks 表包含为 house_id 标识的每个房屋安排的任务列表。 task_status 表跟踪每个任务的进度，包括描述及其当前状态。

房子任务：

task_id (int, primary key): A unique identifier for each task.
house_id (int): The identifier of the house for which the task is scheduled.
task_name (varchar): The name of the task to be performed.

任务状态：

id (int, primary key): A unique identifier for each task status record.
task_id (int): The identifier of the task which this status relates to.
description (varchar): A brief description of the task status.
task_status (varchar): The current status of the task; it can be 'Completed', 'In Progress', or NULL if the status has not been set.

编写一个 SQL 查询，为每个房屋生成报告。报告应包括以下列：

house_id: The identifier of the house.
total_tasks: The total number of tasks scheduled for the house.
completed_tasks: The number of tasks that have been completed. A task is considered completed if all of its corresponding statuses in the task_status table are set to 'Completed'.
incomplete_tasks: The number of tasks that are not completed. A task is considered incomplete if any of its statuses are not 'Completed' or if the task does not have any status record in the task_status table.
The result should be ordered by house_id in descending order.

对于此示例数据：

房子任务：

| task_id | house_id | task_name |
+---------+----------+-----------+
| 1       | 1        | Paint     |
| 2       | 1        | Plumb     |
| 3       | 1        | Garden    |
| 4       | 2        | Electric  |
| 5       | 2        | Roof      |
| 6       | 3        | Cleanup   |
| 7       | 4        | Extra Work|

任务状态：

| task_id | description | task_status |
+---------+-------------+-------------+
| 1       | Desc 1      | Completed   |
| 1       | Desc 2      | Completed   |
| 1       | Desc 3      | Completed   |
| 2       | Desc 4      | In Progress |
| 2       | Desc 5      | In Progress |
| 3       | Desc 6      | In Progress |
| 3       | Desc 7      | Completed   |
| 4       | Desc 8      | Completed   |
| 5       | Desc 9      | NULL        |
| 7       | Desc 10     | NULL        |
| 7       | Desc 11     | Completed   |

所需的输出如下：

| house_id | total_tasks | completed_tasks | incomplete_tasks |
+----------+-------------+-----------------+------------------+
| 4        | 1           | 0               | 1                |
| 3        | 1           | 0               | 1                |
| 2        | 2           | 1               | 1                |
| 1        | 3           | 1               | 2                |

这是我的查询，不计算正确完成的任务：

SELECT
    house_id, 
    COUNT(DISTINCT house_tasks.task_id) AS total_tasks,
    COUNT(DISTINCT CASE WHEN task_status.task_status = 'Completed' THEN house_tasks.task_id END) AS completed_tasks,
    COUNT(DISTINCT CASE WHEN task_status.task_status IS NULL OR task_status.task_status <> 'Completed' THEN house_tasks.task_id END) AS incomplete_tasks
FROM 
    house_tasks
LEFT JOIN 
    task_status ON house_tasks.task_id = task_status.task_id
GROUP BY 
    house_id 
ORDER BY 
    house_id DESC;
    

我的查询显示：

house_id    total_tasks completed_tasks incomplete_tasks
4           1           1                1
3           1           0                1
2           2           1                1
1           3           1                2 

Answer 1

如评论中所述，似乎每个任务都有多行，每次状态更新时都有一行。

假设任务完成后无法重新打开，计算未完成任务的最简单方法是计算已完成的任务（就像您所做的那样），然后从任务总数中减去它：

SELECT   house_id, 
         total_tasks,
         completed_tasks
         total_tasks - completed_tasks AS incomplete_tasks
FROM     (SELECT    house_id, 
                    COUNT(DISTINCT house_tasks.task_id) AS total_tasks,
                    COUNT(DISTINCT CASE WHEN task_status.task_status = 'Completed'     
                                        THEN house_tasks.task_id
                                   END) AS completed_tasks,
          FROM      house_tasks
          LEFT JOIN task_status ON house_tasks.task_id = task_status.task_id
          GROUP BY  house_id) t
ORDER BY house_id DESC;