Агрегирование Несколько столбцов в SQL

Question

Предположим, у меня есть таблица, которая выглядит следующим образом

id | location | dateHired | dateRehired | dateTerminated 
1 | 1  | 10/1/2011 | NULL  | 12/1/2011 
2 | 1  | 10/3/2011 | 11/1/2011 | 12/31/2011 
3 | 5  | 10/5/2011 | NULL  | NULL 
4 | 5  | 10/5/2011 | NULL  | NULL 
5 | 7  | 11/5/2011 | NULL  | 12/1/2011 
6 | 10  | 11/2/2011 | NULL  | NULL 

, и я хотел, чтобы уплотнить, что в сводной таблице, такие, что:

location | date  | hires | rehires | terms 
1  | 10/1/2011 | 1 | 0 |  0 
1  | 10/3/2011 | 1 | 0 |  0 
1  | 11/1/2011 | 0 | 1 |  0 
1  | 12/1/2011 | 0 | 0 |  1 
1  | 12/31/2011 | 1 | 0 |  0 
5  | 10/5/2011 | 2 | 0 |  0 

т.д.

- как бы выглядел SQL? Я думал, что это было бы чем-то вроде:

SELECT 
    e.location 
    , -- ? 
    ,SUM(CASE WHEN e.dateHired IS NOT NULL THEN 1 ELSE 0 END) AS Hires 
    ,SUM(CASE WHEN e.dateRehired IS NOT NULL THEN 1 ELSE 0 END) As Rehires 
    ,SUM(CASE WHEN e.dateTerminated IS NOT NULL THEN 1 ELSE 0 END) As Terms 
FROM 
    Employment e 
GROUP BY 
    e.Location 
    ,--? 

Но я не очень заинтересован, если это абсолютно правильно или нет?

EDIT - это для SQL 2008 R2.

Кроме того,

INNER JOIN на сегодняшний день столбцов предполагает, что есть значения для всех трех категорий, что ложно; которая является первоначальной проблемой, которую я пытался решить. Я думал о чем-то вроде COALESCE, но это тоже не имеет смысла.

Answer 1

Я уверен, что есть, вероятно, более простой и элегантный способ решить эту проблему. Тем не менее, это самый простой, быстрый, что я могу думать об этом поздно, что работает.

CREATE TABLE #Temp 
(
    Location INT, 
    Date DATETIME, 
    HireCount INT, 
    RehireCount INT, 
    DateTerminatedCount INT 
) 

--This will keep us from having to do an insert if does not already exist 
INSERT INTO #Temp (Location, Date) 
SELECT DISTINCT Location, DateHired FROM Employment 
UNION 
SELECT DISTINCT Location, DateRehired FROM Employment 
UNION 
SELECT DISTINCT Location, DateTerminated FROM Employment 

UPDATE #Temp 
SET HireCount = Hired.HireCount 
FROM #Temp 
JOIN 
(
    SELECT Location, DateHired AS Date, SUM(*) AS HireCount 
    FROM Employment 
    GROUP BY Location, DateHired 
) AS Hired 

UPDATE #Temp 
SET RehireCount= Rehire.RehireCount 
FROM #Temp 
JOIN 
(
    SELECT Location, DateRehired AS Date, SUM(*) AS RehireCount 
    FROM Employment 
    GROUP BY Location, DateRehired 
) AS Rehire 
    ON Rehire.Location = #Temp.Location AND Rehire.Date = #Temp.Date 

UPDATE #Temp 
SET DateTerminatedCount = Terminated.DateTerminatedCount 
FROM #Temp 
JOIN 
(
    SELECT Location, DateTerminated AS Date, SUM(*) AS DateTerminatedCount 
    FROM Employment 
    GROUP BY Location, DateTerminated 
) AS Terminated 
    ON Terminated.Location = #Temp.Location AND Terminated.Date = #Temp.Date 

SELECT * FROM #Temp 

Answer 2

SELECT * FROM 
(SELECT location, dateHired as date, COUNT(1) as hires FROM mytable GROUP BY location, date) H 
INNER JOIN 
(SELECT location, dateReHired as date, COUNT(1) as rehires FROM mytable GROUP BY location, date) R ON H.location = R.location AND H.dateHired = R.dateRehired 
INNER JOIN 
(SELECT location, dateTerminated as date, COUNT(1) as terminated FROM mytable GROUP BY location, date) T 
ON H.location = T.location AND H.dateHired = T.dateTerminated 

Answer 3

Как о чем-то вроде:

with dates as (
    select distinct location, d from (
     select location, dateHired as [d] 
     from tbl 
     where dateHired is not null 

     union all 

     select location, dateRehired 
     from tbl 
     where dateRehired is not null 

     union all 

     select location, dateTerminated 
     from tbl 
     where dateTerminated is not null 
    ) 
) 

select location, [d], 
    (
     select count(*) 
     from tbl 
     where location = dates.location 
      and dateHired = dates.[d] 
    ) as hires, 
    (
     select count(*) 
     from tbl 
     where location = dates.location 
      and dateRehired = dates.[d] 
    ) as rehires, 
    (
     select count(*) 
     from tbl 
     where location = dates.location 
      and dateTerminated = dates.[d] 
    ) as terms 
from dates 

не имеют SQL сервер под рукой, или я бы проверить его.