Question

比方说，我有3个表：约会，医生和部门。

约会：

id     status      doctor_id
----   ---------   ---------
1      approved    1
2      cancelled   4
3      approved    4
4      approved    1
5      approved    4
6      NULL        5
7      approved    2
8      NULL        5
9      approved    4
10     approved    3
11     cancelled   1
12     NULL        4
13     approved    3
14     cancelled   1
15     approved    4
16     cancelled   4
17     cancelled   2
18     NULL        4
19     cancelled   1
20     cancelled   4

医生：

id     name        department_id
----   ---------   -------------
1      John        1
2      Robert      2
3      Patricia    3
4      Mary        1
5      Susan       3

部门：

id     name
----   ---------
1      Dermatology
2      Neurology
3      Radiology

我需要的是经过医生分组的皮肤科批准和取消（批准/（已批准+已取消）* 100）总数的批准预约百分比。

我使用了以下查询，这使我更接近解决方案。

SELECT COUNT(*) AS appointment_count,
       doctors.name AS doctor_name,
       appointments.status AS appointment_status
FROM appointments
LEFT JOIN doctors ON appointments.doctor_id = doctors.id
LEFT JOIN departments ON doctors.department_id = departments_id
WHERE departments.id = 1
GROUP BY doctors.id,
         appointments.status

结果：

count    doctor_name    appointment_status
-----    -----------    ---------
2        John           approved
3        John           cancelled
0        John           NULL
4        Mary           approved
3        Mary           cancelled
2        Mary           NULL

但我需要每位医生批准/（已批准+已取消）的百分比。所以结果应该是：

approved_percentage   doctor_name
-------------------   -----------
%40                   John
%57                   Mary

我怎样才能达到这样的效果？

Answer 1

您只能由医生进行分组，然后使用count（if（））这样的功能：

SELECT d.name, count(*), count(if(a.status = 'approved', 1, null)) approved_count,  
    count(if(a.status = 'approved', 1, null))/count(*) * 100 approved_percentage
    FROM doctors d
    INNER JOIN appointments a ON a.doctor_id = d.id
    GROUP BY d.name

Answer 2

在MySQL中，进行条件计数和平均值的最简单方法就是使用布尔表达式。这表明：

SELECT d.name, count(*),
       SUM(a.status = 'approved') as approved_count,  
       AVG(a.status = 'approved') * 100 as approved_percentage
FROM doctors d INNER JOIN
     appointments a
     ON a.doctor_id = d.id
GROUP BY d.name;

MySQL BY查询中的值百分比

2 个答案: