Question

我有一些看起来像这样的期间数据：

PName        DtFrom       DtTo          Amount
Period_1     2018-01-01   2018-01-10    100
Period_2     2018-01-03   2018-01-08    10
Period_3     2018-01-05   2018-01-12    1

我想获得以下时间序列，添加每个时期日期框架内的金额：

2018-01-01   100
2018-01-02   100
2018-01-03   110
2018-01-04   110
2018-01-05   111
2018-01-06   111
2018-01-07   111
2018-01-08   111
2018-01-09   101
2018-01-10   101
2018-01-11   1
2018-01-12   1

我使用DATE_ADD，DATEDIFF等进行了大量研究，也使用了其他StackOverflow问题。但没有成功。有什么想法吗？

Answer 1

虽然我真的主张在应用程序代码中处理这个问题，但这是一个使用一个小整数实用程序表的解决方案。您可以使用日历表，或者即时构建一个整数序列...

DROP TABLE IF EXISTS my_table;

CREATE TABLE my_table
(Perid SERIAL PRIMARY KEY
,DtFrom DATE NOT NULL
,DtTo DATE NOT NULL          
,Amount INT NOT NULL
);

INSERT INTO my_table VALUES
(1,'2018-01-01','2018-01-10',100),
(2,'2018-01-03','2018-01-08',10),
(3,'2018-01-05','2018-01-12',1);

SELECT * FROM ints;
+---+
| i |
+---+
| 0 |
| 1 |
| 2 |
| 3 |
| 4 |
| 5 |
| 6 |
| 7 |
| 8 |
| 9 |
+---+

SELECT dtfrom + INTERVAL y.i DAY dt
     , SUM(x.amount) total 
  FROM my_table x 
  JOIN ints y 
 WHERE dtfrom + INTERVAL y.i DAY <= dtto 
 GROUP 
    BY dt;
+------------+-------+
| dt         | total |
+------------+-------+
| 2018-01-01 |   100 |
| 2018-01-02 |   100 |
| 2018-01-03 |   110 |
| 2018-01-04 |   110 |
| 2018-01-05 |   111 |
| 2018-01-06 |   111 |
| 2018-01-07 |   111 |
| 2018-01-08 |   111 |
| 2018-01-09 |   101 |
| 2018-01-10 |   101 |
| 2018-01-11 |     1 |
| 2018-01-12 |     1 |
+------------+-------+

如果dtfrom和dtto之间的差距可能超过10天，您可以按照以下方式扩展解决方案......

SELECT dtfrom + INTERVAL i2.i*10 + i1.i DAY dt
     , SUM(x.amount) total
  FROM my_table x
  JOIN ints i1
  JOIN ints i2
 WHERE dtfrom + INTERVAL i2.i*10 + i1.i DAY <= dtto
 GROUP
    BY dt;

Answer 2

试试这个

select
      sum(if(STR_TO_DATE(?, '%d-%m-%Y') >= so_period_date.dt_from and 
             STR_TO_DATE(?, '%d-%m-%Y') <= so_period_date.dt_to,
          so_period_date.amount, 0)) as amount
from so_period_date

Answer 3

试试这个：

SELECT A.`DATE`, B.Amount
FROM 
(SELECT TO_DATE('2018-01-01','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-02','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-03','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-04','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-05','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-06','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-07','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-08','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-09','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-10','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-11','YYYY-MM-DD') `DATE` UNION ALL
 SELECT TO_DATE('2018-01-12','YYYY-MM-DD') `DATE`  
) A LEFT JOIN yourTable B ON A.`DATE` BETWEEN B.DtFrom AND b.DtTo;

如果您必须在sql中执行此查询，那么它很容易在应用程序逻辑中获得预期结果，就像@Strawberry在他对您的问题的评论中指出的那样。

创建从入口到入口的时间序列

3 个答案: