Question

我无法弄清楚如何将数据字段与日期相加...

我的查询当前输出与此类似的内容：

Date    Count
12/1    3
12/2    5
12/3    2

我想添加一个总结count列的列：

Date    Count    To-Date
12/1    3        3
12/2    5        8
12/3    2        10

有什么想法吗？

***更新 - 我正在尝试执行此操作的表格稍微复杂一些：

ID  Date    Count for Date
X   10/11/11    10  
X   10/11/11    10  
X   14/11/11    2   
X   14/11/11    2   
X   22/11/11    21  
X   23/11/11    50
X   23/11/11    50

使用此代码：

Sum(Count for Date) over (partition by x, date order by date rows between unbounded preceding and current row)

我最终得到了：

Answer 1

[更新以反映这一变化，我猜测结果应该是什么，我猜你想加入当天？如果没有，你还可以显示你的预期结果吗？]

分析函数可以执行此操作：

CREATE TABLE test(id int, count int);

INSERT INTO test VALUES(10, 10);
INSERT INTO test VALUES(10, 10);
INSERT INTO test VALUES(14, 2);
INSERT INTO test VALUES(14, 2);
INSERT INTO test VALUES(22, 21);
INSERT INTO test VALUES(23, 50);
INSERT INTO test VALUES(23, 50);

SELECT id, SUM(count) AS sum_for_day
  FROM test
  GROUP BY id
  ORDER BY id;

 id | sum_for_day
----+-------------
 10 |          20
 14 |           4
 22 |          21
 23 |         100
(4 rows)

SELECT id, sum_for_day
      ,sum(sum_for_day) over (order by id rows between unbounded preceding and current row) AS to_date
  FROM ( SELECT id, SUM(count) AS sum_for_day
       FROM test
           GROUP BY id
           ORDER BY id) x;

 id | sum_for_day | to_date
----+-------------+---------
 10 |          20 |      20
 14 |           4 |      24
 22 |          21 |      45
 23 |         100 |     145
(4 rows)

Answer 2

假设您希望每个不同的ID的总数不同，那么您需要类似

的内容

SELECT id, 
       date_column, 
       count_for_date, 
       sum(count_for_date) over (partition by id order by date_column) running_total
  FROM some_table_name

如果您想忽略ID并按date_column汇总所有行，只需删除分析函数中的partition by

SELECT id, 
       date_column, 
       count_for_date, 
       sum(count_for_date) over (order by date_column) running_total
  FROM some_table_name

SQL - 迄今为止的和数据字段

2 个答案: