按日期汇总多个记录 [英] Aggregating multiple records by date

问题描述

我有一个看起来像这样的模型:

I have a Model that looks like:

MedicationAdherence {
                 :id => :integer,
     :adherence_date => :date,
     :scheduled_time => :string,
    :acknowledged_at => :datetime,
         :patient_id => :integer,
         :created_at => :datetime,
         :updated_at => :datetime
}

我有7条记录(与patient_id相同):

I have 7 records (same patient_id):

{ id: 1, adherence_date: 2017-10-01, scheduled_time: 'morning', acknowledged_at: Tue, 31 Oct 2017 19:59:19 UTC +00:00 }
{ id: 2, adherence_date: 2017-10-01, scheduled_time: 'afternoon', acknowledged_at: nil }
{ id: 3, adherence_date: 2017-10-01, scheduled_time: 'night', acknowledged_at: Tue, 31 Oct 2017 19:59:19 UTC +00:00 }
{ id: 4, adherence_date: 2017-10-02, scheduled_time: 'morning', acknowledged_at: Tue, 31 Oct 2017 19:59:19 UTC +00:00 }
{ id: 5, adherence_date: 2017-10-02, scheduled_time: 'afternoon', acknowledged_at: Tue, 31 Oct 2017 19:59:19 UTC +00:00 }
{ id: 6, adherence_date: 2017-10-02, scheduled_time: 'evening', acknowledged_at: Tue, 31 Oct 2017 19:59:19 UTC +00:00 }
{ id: 7, adherence_date: 2017-10-02, scheduled_time: 'night', acknowledged_at: nil }

我想要的结果是将上面的记录分组为以下输出:

My desired outcome is to group the records above into the following output:

{
    "adherence_date" => 2017-10-1,
           "morning" => 1,
         "afternoon" => 0,
           "evening" => nil,
             "night" => 1
},
{
    "adherence_date" => 2017-10-2,
           "morning" => 1,
         "afternoon" => 1,
           "evening" => 1,
             "night" => 0
}

如果没有记录(2017年10月1日晚上)，则应返回nil.当有记录但没有Confirmed_at时，它应该返回false(0)，而当有Confirmed_at时，它返回true(1)

When there is no-record (evening 2017-10-1) it should return nil. When there is a record but no acknowledged_at it should return false (0), and when there is acknowledged_at returns true(1)

下面是我用来尝试合并所有这些数据的查询，但是它给了我重复的记录.如何将我的数据汇总到上面的数据中...我敢肯定有一种更简单的方法可以做到这一点

Below is the query I used to try and combine all this data, but it gives me duplicate records. How can I sum my data into what I have above...I'm sure there's a simplier way to do this

WITH
  adherences AS (
    SELECT * FROM medication_adherences WHERE patient_id = 10049
  ),

  morning AS (
    SELECT adherence_date,
      CASE acknowledged_at WHEN null THEN 0 ELSE 1 END as morning
    FROM adherences
    WHERE scheduled_time = 'morning'
  ),

  afternoon as (
    SELECT adherence_date,
      CASE acknowledged_at WHEN null THEN 0 ELSE 1 END as afternoon
    FROM adherences
    WHERE scheduled_time = 'afternoon'
  ),

  evening as (
    SELECT adherence_date,
      CASE acknowledged_at WHEN null THEN 0 ELSE 1 END as evening
    FROM adherences
    WHERE scheduled_time = 'evening'
  ),

  night as (
    SELECT adherence_date,
      CASE acknowledged_at WHEN null THEN 0 ELSE 1 END as night
    FROM adherences
    WHERE scheduled_time = 'night'
  )

SELECT morning.morning, afternoon.afternoon, evening.evening, night.night, adherences.adherence_date
FROM adherences
LEFT JOIN morning ON morning.adherence_date = adherences.adherence_date
LEFT JOIN afternoon ON afternoon.adherence_date = adherences.adherence_date
LEFT JOIN evening ON evening.adherence_date = adherences.adherence_date
LEFT JOIN night ON night.adherence_date = adherences.adherence_date

我正在运行oracle-12c

编辑

好像我必须在查询中添加GROUP BY morning.morning, afternoon.afternoon, evening.evening, night.night, adherences.adherence_date才能正确分组.有没有更简单的方法来汇总这些数据?

Looks like I had to add GROUP BY morning.morning, afternoon.afternoon, evening.evening, night.night, adherences.adherence_date to my query for it to properly group by. Is there a simpler way to aggregate this data?

推荐答案

我假设(patient_id, adherence_date, scheduled_time)在您的表格中是唯一的，这意味着患者可以在每个时段"和日期中预订一次.

I'm assuming that (patient_id, adherence_date, scheduled_time) is unique in your table, meaning that a patient can book once per "slot" and date.

with medication_adherences  as(
-- This is your test data
   select 10049 as patient_id, 1 as id, date '2017-10-01' as adherence_date, 'morning'    as scheduled_time, timestamp '2017-10-31 19:59:19' as acknowledged_at from dual union all
   select 10049 as patient_id, 2 as id, date '2017-10-01' as adherence_date, 'afternoon'  as scheduled_time, null                            as acknowledged_at from dual union all                          
   select 10049 as patient_id, 3 as id, date '2017-10-01' as adherence_date, 'night'      as scheduled_time, timestamp '2017-10-31 19:59:19' as acknowledged_at from dual union all
   select 10049 as patient_id, 4 as id, date '2017-10-02' as adherence_date, 'morning'    as scheduled_time, timestamp '2017-10-31 19:59:19' as acknowledged_at from dual union all
   select 10049 as patient_id, 5 as id, date '2017-10-02' as adherence_date, 'afternoon'  as scheduled_time, timestamp '2017-10-31 19:59:19' as acknowledged_at from dual union all
   select 10049 as patient_id, 6 as id, date '2017-10-02' as adherence_date, 'evening'    as scheduled_time, timestamp '2017-10-31 19:59:19' as acknowledged_at from dual union all
   select 10049 as patient_id, 7 as id, date '2017-10-02' as adherence_date, 'night'      as scheduled_time, null                            as acknowledged_at from dual
)
select adherence_date
      ,sum(case when scheduled_time = 'morning'   then nvl2(acknowledged_at,1,0) end) as morning
      ,sum(case when scheduled_time = 'afternoon' then nvl2(acknowledged_at,1,0) end) as afternoon
      ,sum(case when scheduled_time = 'evening'   then nvl2(acknowledged_at,1,0) end) as evening
      ,sum(case when scheduled_time = 'night'     then nvl2(acknowledged_at,1,0) end) as night
  from medication_adherences 
 where patient_id = 10049
 group
    by adherence_date;

逻辑原理如下:

• 如果accepted_at为null，则我们将汇总0(通过nvl2)
• 如果accepted_at为不为空，则我们汇总1(通过nvl2)
• 如果此时隙没有记录，我们将合计null(因为...失败的情况)

• if acknowledged_at is null then we aggregate 0 (via nvl2)
• if acknowledged_at is not null then we aggregate 1 (via nvl2)
• if there is no record for this timeslot we aggregate null (since case when ... failed)

这篇关于按日期汇总多个记录的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！