Question

我是SQL的新手，请原谅我这个天真的问题。

我的日期范围为“ 20180903-20180905”，因此距3rd Sep to 5th Sep 3天。

现在，每天我们有24个军事小时，从0-23开始，我们为a range of 4 at a time in a single day写一个查询。

所以，假设，

select 
COUNT(wk_id),
SUM(total_occurances) as total_wk_occurances,
SUM(SUCCEEDED_instances) as total_successful_occurances,
SUM(FAILED_instances) as total_error_occurances 
from
(
    select w1.wk_id,
    COUNT(w1.wk_occurance_id) as total_occurances,
    sum(case when w1.status='SUCCEEDED' then 1 else 0 end ) as SUCCEEDED_instances, 
    sum(case when w1.status !='SUCCEEDED' then 1 else 0 end ) as FAILED_instances  
    from 
        work_instances w1 
    inner join  
        time_table td2 
   on  
        w1.end_time = td2.time_id
   where  
        (td2.military_hour between 0 and 3 and end_date='20180903') group by w1.wf_id
) as sub_q1

现在，我在20180903个军事小时范围0-3中进行了测试。然后返回类似

的行

46 | 224 | 208 | 16

我是否必须再用4-7, 8-11, 12- 15, 16-19, 20 - 23之类的范围再写5次相同的查询？然后汇总并返回具有6行的查询集？

如何更好地编写它？

此外，它仅适用于20180903。

如果其他2 days and return all 18 rows each(6 each day)要怎么做？

例如

结果集可能看起来像

COUNT   |   total_wk_occurances  |  total_succeessful_occurances  |  total_error_Occurances

46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25
46      |   224                  |  208                          | 16
34      |   100                  |  75                           | 25

Answer 1

使用条件聚合和group by。像这样：

select end_date, 
       (floor(td2.military_hour / 4) * 4) as military_hour,
       count(w1.wk_occurance_id) as total_occurances,
       sum( (w1.status = 'SUCCEEDED')::int ) as SUCCEEDED_instances, 
       sum( (w1.status <> 'SUCCEEDED')::int ) as FAILED_instances  
from work_instances w1 inner join  
     time_table td2 
     on   w1.end_time = td2.time_id
where end_date >= 20180901' and end_date <2018
group by end_date, (floor(td2.military_hour / 4) * 4)

在PostgreSQL的单个结果集中组合多个相同类型的SQL查询

1 个答案: