2つの選択カウントステートメントの結合方法

Question

たとえば、学生がアクティビティを行うたびにテーブル'studentActivities'が記録されているテーブルがあるとします。これと同じように：

activity   countryofbirth name event_date 
School Swimming USA    Bob  5/21/2017 12:50 
Park Swimming Australia   Sarah 2/11/2017 19:50 
Park Swimming Australia   Sarah 2/13/2017 16:50 
Park Running  USA    Bob  2/10/2017 11:50 
School Tennis USA    Bob  2/12/2017 11:50 
NULL    USA    Jane 8/4/2016 13:30 

私はそれがIn-School ActvityかOut-of-School Activityであれば、学生が基づいん数の異なる活動をカウントしようとしています。

Student_Name Country_of_Birth In-School_Activities Out_of_School_Activities 
Bob   USA    2     1 
Sarah  Australia  0     1 
Jane   USA    0     0 

私が試してみました：

SELECT 
studentActivities.name AS [student_name], 
studentActivities.countryofbirth AS County_of_Birth], 
COUNT (DISTINCT activity) as [In-School Activities] 

FROM studentActivities 

WHERE studentActivities.activity IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 

GROUP BY studentActivities.name, studentActivities.countryofbirth 

UNION 

SELECT 
studentActivities.name AS [student_name], 
studentActivities.countryofbirth AS [County_of_Birth], 
COUNT (DISTINCT activity) as [Out_of_School Activities] 

FROM studentActivities 

WHERE studentActivities.activity NOT IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 

GROUP BY studentActivities.name, studentActivities.countryofbirth 

School Swimming, School Running, School Soccer, School Tennis

は、私は何を取得したいと思いするほどのテーブルです：

この

は経由 4で、学校の活動によって区別されます

しかし、これは私が望む結果を私に与えることはありません。私が望む結果を得るにはどうすればいいですか？

Answer 1

条件付き集約を使用してcount(distinct ...)

select 
    Name 
    , CountryOfBirth 
    , InSchoolActivies = count(distinct case when activity in ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') then activity end) 
    , OutOfSchoolActivies = count(distinct case when activity not in ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') then activity end) 
from StudentActivites 
group by Name, CountryOfBirth 

rextesterデモで：http://rextester.com/EMUWU73595

返品：（サンプルデータでBobの生年月日を修正した後）

+-------+----------------+------------------+---------------------+ 
| Name | CountryOfBirth | InSchoolActivies | OutOfSchoolActivies | 
+-------+----------------+------------------+---------------------+ 
| Bob | USA   |    2 |     1 | 
| Jane | USA   |    0 |     0 | 
| Sarah | Australia  |    0 |     1 | 
+-------+----------------+------------------+---------------------+ 

Answer 2

あなたは

WITH InSchool AS (
SELECT 
studentActivities.name AS [student_name], 
studentActivities.countryofbirth , 
COUNT (DISTINCT activity) as [In-School Activities] 
FROM studentActivities 
WHERE studentActivities.activity IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 
GROUP BY studentActivities.name, studentActivities.countryofbirth 
) 
, OutOfSchool AS ( 
SELECT 
studentActivities.name AS [student_name], 
studentActivities.countryofbirth, 
COUNT (DISTINCT activity) as [Out_of_School Activities] 
FROM studentActivities 
WHERE studentActivities.activity NOT IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 
GROUP BY studentActivities.name, studentActivities.countryofbirth 
) 
SELECT 
    COALESCE(InSchool.student_name, OutOfSchool.student_name) [Student_Name], 
    COALESCE(InSchool.CountryOfBirth, OutOfSchool.CountryOfBirth) CountryOfBirth, 
    COALESCE(InSchool.[In-School Activities],0) [In-School Activities], 
    COALESCE(OutOfSchool.[Out_of_School Activities],0) [Out_of_School Activities] 
FROM InSchool 
FULL OUTER JOIN OutOfSchool 
ON InSchool.[student_name] = OutOfSchool.[Student_name] 
AND InSchool.CountryOfBirth = OutOfSchool.CountryOfBirth 

Answer 3

に参加し、この私の心に来るもの、私は確認していなかったが、これは、私はあなたが参加を使用する必要があるタスク

SELECT 
name AS [student_name], 
countryofbirth AS [Manufacturer], 
sum (activity IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis')) as [In-School Activities], 
sum (activity NOT IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis')) as [Out_of_School Activities] 
FROM (select distinct studentActivities.name, studentActivities.countryofbirth, activity) x 
GROUP BY name, countryofbirth; 

Answer 4

を得た方法で完全外部を使用する必要があります私はCaseステートメントを使用します

select a.[student_name], a.[student_name], a.[In-School Activities], b.[Out_of_School Activities] 
from ( 
    SELECT 
    studentActivities.name AS [student_name], 
    studentActivities.countryofbirth AS [Manufacturer], 
    COUNT (DISTINCT activity) as [In-School Activities] 
    FROM studentActivities 
    WHERE studentActivities.activity IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 
    GROUP BY studentActivities.name, studentActivities.countryofbirth 
) a 
left join (
    SELECT 
    studentActivities.name AS [student_name], 
    studentActivities.countryofbirth AS [Manufacturer], 
    COUNT (DISTINCT activity) as [Out_of_School Activities] 
    FROM studentActivities 
    WHERE studentActivities.activity NOT IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 
    GROUP BY studentActivities.name, studentActivities.countryofbirth 
) b on a.[student_name] = b.[student_name] and a.[Manufacturer] = b.[Manufacturer] 

Answer 5

（とない組合）：

SELECT 
Sa.name AS [student_name], 
Sa.countryofbirth, 

Sum(Case when sa.activity IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 
then 1 else 0 end) as In_school_Activities, 

Sum(Case when sa.activity IN ('School Swimming', 'School Running', 'School Soccer', 'School Tennis') 
then 0 else 1 end) as Out_of_school_Activities 


FROM (select distinct student_name, country _of_birth, activity from studentActivities) sa 

GROUP BY sa.name, sa.countryofbirth 

重要なアプリケーションでは、名前は一意であるとは限りません。あなたが識別子を持っているなら（そしてあなたが名前を持っている時はいつでも）、それをグループ化するほうがいいです。

Answer 6

ピボットがこれを解決します。私はあなたが提供したデータでそれをテストし、それは期待通りに機能しました。

SELECT 
    Name 
    ,countryofbirth 
    , [In-School Activity] 
    ,[Out-of-School Activity] 
FROM(
    SELECT 
     name 
     ,countryofbirth 
     ,CASE Activity 
      WHEN 'School Swimming' THEN 'In-School Activity' 
      WHEN 'School Running' THEN 'In-School Activity' 
      WHEN 'School Soccer' THEN 'In-School Activity' 
      WHEN 'School Tennis' THEN 'In-School Activity'   
      ELSE 'Out-of-School Activity' 
     END class 
    FROM studentactivities 
    WHERE Activity IS NOT NULL 
)AS base 
PIVOT(
    COUNT (CLASS) 
    FOR CLASS IN ([In-School Activity],[Out-of-School Activity]) 
)AS Pivoted 

Answer 7

ボブには2つの生殖拠点があります。それは2つの別々のボブですか？それともエラーですか？それは100％正確ではないかもしれませんので、私はこれをテストしていないが、私は、共通テーブル式を好む：

;WITH cte 
AS (
SELECT DISTINCT NAME AS [student_name] 
    ,countryofbirth 
    ,activity 
    ,CASE 
     WHEN activity IN (
       'School Swimming' 
       ,'School Running' 
       ,'School Soccer' 
       ,'School Tennis' 
       ) 
      THEN 1 
     ELSE 0 
     END AS [In-school_activity_count] 
    ,CASE 
     WHEN activity IS NOT NULL 
      AND activity NOT IN (
       'School Swimming' 
       ,'School Running' 
       ,'School Soccer' 
       ,'School Tennis' 
       ) 
      THEN 1 
     ELSE 0 
     END AS [Out-of-school_activity_count] 
FROM studentActivities 
) 
SELECT student_name 
,countryofbirth 
,SUM([In-school_activity_count]) AS [In-School_Activities] 
,SUM([Out-of-school_activity_count]) AS [Out_of_School_Activities] 
FROM cte 
GROUP BY student_name 
,countryofbirth 

Answer 8

以下は簡単で、動作します：

select name,countryofbirth,sum(schoolflag) as 
In_School_Activity,sum(nonschool) Out_Of_School_Activity 
from 
(
select activity,name,countryofbirth, 
case when activity like 'School%' then 1 else 0 end as schoolflag , 
case when activity not like 'School%' then 1 else 0 end as nonschool 
from schoolactivity 
)a 
group by name,countryofbirth