mysql的計算天

Question

我有一個表，像這樣的命令：

+----------+-------------+------------+ 
| order_id | customer_id | order_date | 
+----------+-------------+------------+ 
| 404 |  16  | 2006-11-04 | 
| 726 |  16  | 2007-11-01 | 
| 440 |  16  | 2010-03-15 | 
| 515 |  16  | 2011-04-29 | 
| 539 |  16  | 2011-11-01 | 
| 293 |  89  | 2001-08-28 | 
| 156 |  89  | 2002-05-03 | 
| 238 |  89  | 2003-09-15 | 
| 299 |  28  | 2001-10-26 | 
| 314 |  28  | 2004-03-18 | 
| 142 |  79  | 2002-01-21 | 
| 313 |  12  | 2004-03-11 | 
| 388 |  12  | 2005-11-22 | 
| 507 |  12  | 2007-03-27 | 
| 286 |  14  | 2006-04-17 | 
| 200 |  14  | 2008-05-25 | 
| 130 |  14  | 2009-06-24 | 
| 527 |  19  | 2011-05-14 | 
| 298 |  76  | 2001-10-20 | 
| 165 |  76  | 2002-07-27 | 
| 212 |  11  | 2003-02-17 | 
| 412 |  11  | 2006-08-02 | 
| 423 |  11  | 2007-03-28 | 
| 388 |  10  | 2005-01-07 | 
| 390 |  65  | 2006-06-07 | 
| 44 |  65  | 2008-01-07 | 
| 269 |  16  | 2008-02-25 | 
| 413 |  14  | 2007-12-01 | 
| 675 |  14  | 2007-09-11 | 
| 260 |  14  | 2009-10-20 | 
+----------+-------------+------------+ 

現在我需要一個查詢至極給我這樣一個列表：

+-------------+------------+------------+------+ 
| customer_id | date1 | date2 | days | 
+-------------+------------+------------+------+ 
|  16  | 2006-11-04 | 2007-11-01 | 362 | 
|  16  | 2007-11-01 | 2010-03-15 | 865 | 
|  16  | 2010-03-15 | 2011-04-29 | 410 | 
|  16  | 2011-04-29 | 2011-11-01 | 186 | 
|  89  | 2001-08-28 | 2002-05-03 | 248 | 
|  89  | 2002-05-03 | 2003-09-15 | 500 | 
+-------------+------------+------------+------+ 

列日在天時間跨度從一個訂單到每個客戶的下一個訂單。

Answer 1

你可以嘗試這樣的事情：

SELECT A.customer_id, 
    A.date as date1, 
    B.date as date2, 
    datediff(B.date, A.date) as days 
FROM (select @curRowA := @curRowA + 1 AS row_number, 
     customer_id, 
     date from 
     orders 
    ORDER BY customer_id asc, date desc) A 
INNER JOIN (select @curRowB := @curRowB + 1 AS row_number, 
      customer_id, 
      date from orders 
    ORDER BY customer_id asc, date desc offset 1) B 
ON A.row_number = B.row_number AND A.customer_id = B.customer_id 

正如你所看到的，我有一個選擇，從訂單表中的所有記錄，但增加了ROW_NUMBER分爲A子查詢，然後我做同樣的事情，但只是轉移一行向下（偏移量1）並創建B.然後，我只需加入A和B，即可獲得結果。

A 
    row_id, customer, date 
    1, 1, d1 
    2, 1, d2 
    3, 1, d3 
    4, 2, d4 
    5, 2, d5 

B 
    row_id, customer, date 
    1, 1, d2 
    2, 1, d3 
    3, 2, d4 
    4, 2, d5 

當A和B加入

result 
customer, date1, date2 
1, d1, d2 
1, d2, d3 
2, d4, d5 

UPDATE：按照馬克的評論，它可能是更好地嘗試從相關查詢獲取日期2（客戶），您選擇的日期，其體積更小比日期和限制導致1行。

Answer 2

你可以請試試這個，很好的測試...

SELECT distinct(c1.customer_id) as 'customer_id', 

(select min(c2.order_date) from tmp2 c2 where c2.customer_id=c1.customer_id) as date1, (select max(c2.order_date) from tmp2 c2 where c2.customer_id=c1.customer_id) as date2, (DATEDIFF((select max(c40.order_date) from tmp2 c40 where c40.customer_id=c1.customer_id) , (select min(c40.order_date) from tmp2 c40 where c40.customer_id=c1.customer_id))) as 'days' FROM TMP2 as c1 WHERE 1