Подсчитать и повернуть таблицу по дате
Я хотел бы идентифицировать возвращающихся клиентов из таблицы Oracle(11g) следующим образом:
CustID | Date
-------|----------
XC321 | 2016-04-28
AV626 | 2016-05-18
DX970 | 2016-06-23
XC321 | 2016-05-28
XC321 | 2016-06-02
Таким образом, я могу видеть, какие клиенты вернулись в различные окна, например, в течение 10, 20, 30, 40 или 50 дней. Например:
CustID | 10_day | 20_day | 30_day | 40_day | 50_day
-------|--------|--------|--------|--------|--------
XC321 | | | 1 | |
XC321 | | | | 1 |
Я бы даже принял такой результат:
CustID | Date | days_from_last_visit
-------|------------|---------------------
XC321 | 2016-05-28 | 30
XC321 | 2016-06-02 | 5
Я предполагаю, что он использовал бы раздел с помощью оконного предложения с неограниченными следующими и предыдущими предложениями... но я не могу найти подходящих примеров. Есть идеи...? Спасибо
3 ответа
Здесь нет необходимости в оконных функциях, вы можете просто сделать это с помощью условного агрегирования, используя CASE EXPRESSION
:
SELECT t.custID,
COUNT(CASE WHEN (last_visit- t.date) <= 10 THEN 1 END) as 10_day,
COUNT(CASE WHEN (last_visit- t.date) between 11 and 20 THEN 1 END) as 20_day,
COUNT(CASE WHEN (last_visit- t.date) between 21 and 30 THEN 1 END) as 30_day,
.....
FROM (SELECT s.custID,
LEAD(s.date) OVER(PARTITION BY s.custID ORDER BY s.date DESC) as last_visit
FROM YourTable s) t
GROUP BY t.custID
Установка Oracle:
CREATE TABLE customers ( CustID, Activity_Date ) AS
SELECT 'XC321', DATE '2016-04-28' FROM DUAL UNION ALL
SELECT 'AV626', DATE '2016-05-18' FROM DUAL UNION ALL
SELECT 'DX970', DATE '2016-06-23' FROM DUAL UNION ALL
SELECT 'XC321', DATE '2016-05-28' FROM DUAL UNION ALL
SELECT 'XC321', DATE '2016-06-02' FROM DUAL;
Запрос:
SELECT *
FROM (
SELECT CustID,
Activity_Date AS First_Date,
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '10' DAY FOLLOWING )
- 1 AS "10_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '20' DAY FOLLOWING )
- 1 AS "20_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '30' DAY FOLLOWING )
- 1 AS "30_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '40' DAY FOLLOWING )
- 1 AS "40_Day",
COUNT(1) OVER ( PARTITION BY CustID
ORDER BY Activity_Date
RANGE BETWEEN CURRENT ROW AND INTERVAL '50' DAY FOLLOWING )
- 1 AS "50_Day",
ROW_NUMBER() OVER ( PARTITION BY CustID ORDER BY Activity_Date ) AS rn
FROM Customers
)
WHERE rn = 1;
Выход
USTID FIRST_DATE 10_Day 20_Day 30_Day 40_Day 50_Day RN
------ ------------------- ---------- ---------- ---------- ---------- ---------- ----------
AV626 2016-05-18 00:00:00 0 0 0 0 0 1
DX970 2016-06-23 00:00:00 0 0 0 0 0 1
XC321 2016-04-28 00:00:00 0 0 1 2 2 1
Вот ответ, который работает для меня, я основал его на ваших ответах выше, спасибо за вклады от MT0 и Sagi:
SELECT CustID,
visit_date,
Prev_Visit ,
COUNT( CASE WHEN (Days_between_visits) <=10 THEN 1 END) AS "0-10_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 11 AND 20 THEN 1 END) AS "11-20_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 21 AND 30 THEN 1 END) AS "21-30_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 31 AND 40 THEN 1 END) AS "31-40_day" ,
COUNT( CASE WHEN (Days_between_visits) BETWEEN 41 AND 50 THEN 1 END) AS "41-50_day" ,
COUNT( CASE WHEN (Days_between_visits) >50 THEN 1 END) AS "51+_day"
FROM
(SELECT CustID,
visit_date,
Lead(T1.visit_date) over (partition BY T1.CustID order by T1.visit_date DESC) AS Prev_visit,
visit_date - Lead(T1.visit_date) over (
partition BY T1.CustID order by T1.visit_date DESC) AS Days_between_visits
FROM T1
) T2
WHERE Days_between_visits >0
GROUP BY T2.CustID ,
T2.visit_date ,
T2.Prev_visit ,
T2.Days_between_visits;
Это возвращает:
CUSTID | VISIT_DATE | PREV_VISIT | DAYS_BETWEEN_VISIT | 0-10_DAY | 11-20_DAY | 21-30_DAY | 31-40_DAY | 41-50_DAY | 51+DAY
XC321 | 2016-05-28 | 2016-04-28 | 30 | | | 1 | | |
XC321 | 2016-06-02 | 2016-05-28 | 5 | 1 | | | | |