在一個ID內減去多個日期

Question

我試圖在mySQL中完成某些工作，由於缺乏「滯後」功能以及因爲所有關於日期差異的示例，因此很難實現，因爲在所有這些例子總是有一個ID到一個日期。在這裏，我試圖在一個ID內做日期差異，並且當日期差異超過60時，然後返回1的指示符，否則爲0.

不知道最好的方法是怎麼去做這件事。是否將row_number（）與日期結合使用？這個障礙是在多個ID內完成的，因爲我讀過的很多東西都沒有涵蓋。任何方向都會有幫助。 謝謝！

ID | Service Date | Date Difference | Indicator 
1 | 1/22/2016  | 0    | 1 
1 | 3/26/2016  | 64    | 1 
1 | 5/25/2016  | 60    | 0 
1 | 9/15/2016  | 113    | 1 
2 | 8/1/2016  | 0    | 1 
3 | 1/26/2016  | 0    | 1 
3 | 3/9/2016  | 43    | 0 
3 | 4/30/2016  | 52    | 0 
4 | 8/9/2016  | 0    | 1 
5 | 11/19/2016  | 0    | 1 
6 | 10/14/2016  | 0    | 1 
7 | 1/31/2016  | 0    | 1 
7 | 8/11/2016  | 193    | 1 

Answer 1

create view id_and_date as 
select id, service_date from your table; 

create view id_and_date_and_prior as 
select 
a.id, a.service_date, 
coalesce(
    (select max(b.service_date) from id_and_date b 
    where b.id = a.id and b.service_date < a.service_date), 
a.service_date) 
as prior_date 
from id_and_date a 

select a.id, a.service_date, a.prior_date 
date_diff(a.service_date, a.prior_date) as diff, 
case when date_diff(a.service_date, a.prior_date) > 60 
    then 1 else 0 end 
as indicator 
from id_and_date_and_prior a 

Answer 2

可以使用變量，但是這是非常棘手。對於這個可靠地工作，所有的變量都需要在一個單一的表達被分配：

select t.*, datediff(prev_date, date) as diff, 
     (case when datediff(prev_date, date) < 60 then 0 else 1 end) as indicator 
from (select t.*, 
      (case when @id = id 
        then (case when (@prev := @d) = NULL then 'never' -- intentional 
           when (@d := date) = NULL then 'never' -- intentional 
           else @prev 
         end) 
        when (@d := date) = NULL then 'never' -- intentional 
        else NULL 
       end) as prev_date 
     from t cross join 
      (select @id := -1, @d := '') params 
     order by id, date 
    ) t 

Answer 3

發佈，以簡化和糾正功能由@tpdi提供的接聽電話。請接受/ upvote他們的答案，因爲這是非常複製它。

的變化：

• date_diff到DATEDIFF
• 除去創建視圖呼叫贊成t子查詢的
• 分配變量涉及diff值的0初始值
• 指示符0 diff來1
• 當案件有利於IF時更換

SELECT 
c.id, 
c.service_date, 
@diff := DATEDIFF(c.service_date, c.prior_date) AS diff, 
IF(@diff = 0 || @diff > 60, 1, 0) AS indicator 
FROM (
    SELECT 
    a.id, 
    a.service_date, 
    COALESCE(
     (SELECT MAX(b.service_date) 
     FROM t AS b 
     WHERE b.id = a.id 
     AND b.service_date < a.service_date), 
     a.service_date 
    ) AS prior_date 
    FROM t AS a 
) AS c; 

會導致：

| id | service_date | diff | indicator | 
| 1 | 2016-01-22 | 0 | 1   | 
| 1 | 2016-03-26 | 64 | 1   | 
| 1 | 2016-05-25 | 60 | 0   | 
| 1 | 2016-09-15 | 113 | 1   | 
| 2 | 2016-08-01 | 0 | 1   | 
| 3 | 2016-01-26 | 0 | 1   | 
| 3 | 2016-03-09 | 43 | 0   | 
| 3 | 2016-04-30 | 52 | 0   | 
| 4 | 2016-08-09 | 0 | 1   | 
| 5 | 2016-11-19 | 0 | 1   | 
| 6 | 2016-10-14 | 0 | 1   | 
| 7 | 2016-01-31 | 0 | 1   | 
| 7 | 2016-08-11 | 193 | 1   |