获取满足给定条件的连续天数

Question

我的 Oracle 数据库中有以下结构：

Date          Allocation  id
2015-01-01    Same        200
2015-01-02    Good        200
2015-01-03    Same        200
2015-01-04    Same        200
2015-01-05    Same        200
2015-01-06    Good        200

我想要一个查询，它只需要检查前连续几天并获取分配为"Same" 的计数。

我想按日期选择，例如2015-01-05。
示例输出：对于日期2015-01-05，计数为3。

新问题。通过 Lukas Eder 的查询，计数始终为 1 或 2。但预期是3。为什么？！

Date          Allocation  id
2015-01-01    Same        400
2015-01-02    Good        400
2015-01-03    Same        400
2015-01-04    Same        400
2015-01-05    Same        400
2015-01-06    Good        400

来自 Lukas Eder 的代码

 SELECT c
    FROM (
      SELECT allocation, d, count(*) OVER (PARTITION BY allocation, part ORDER BY d) AS c
      FROM (
        SELECT allocation, d,
               d - row_number() OVER (PARTITION BY allocation ORDER BY d) AS part
        FROM t
      )
    )
    WHERE d = DATE '2015-01-05';

预期的输出是这样的，First_day end Last day 不需要：

id   count    first_day   Last_Day
200  3        2015-01-03  2015-01-05
400  3        2015-01-03  2015-01-05

Answer 1

此查询将产生每一行的计数：

SELECT allocation, d, count(*) OVER (PARTITION BY allocation, part ORDER BY d) AS c
FROM (
  SELECT allocation, d,
         d - row_number() OVER (PARTITION BY allocation ORDER BY d) AS part
  FROM t
)
ORDER BY d;

然后您可以对其进行过滤以查找给定行的计数：

SELECT c
FROM (
  SELECT allocation, d, count(*) OVER (PARTITION BY allocation, part ORDER BY d) AS c
  FROM (
    SELECT allocation, d,
           d - row_number() OVER (PARTITION BY allocation ORDER BY d) AS part
    FROM t
  )
)
WHERE d = DATE '2015-01-05';

说明：

派生表用于计算每个日期和分配的不同“分区”part：

  SELECT allocation, d,
         d - row_number() OVER (PARTITION BY allocation ORDER BY d) AS part
  FROM t

结果是：

allocation  d           part
--------------------------------
Same        01.01.15    31.12.14
Good        02.01.15    01.01.15
Same        03.01.15    01.01.15
Same        04.01.15    01.01.15
Same        05.01.15    01.01.15
Good        06.01.15    04.01.15

part 产生的具体日期无关紧要。只是某个日期对于分配中的每个“组”日期都是相同的。然后您可以使用count(*) over(...) 窗口函数计算(allocation, part) 相同值的数量：

SELECT allocation, d, count(*) OVER (PARTITION BY allocation, part ORDER BY d) AS c
FROM (...)
ORDER BY d;

产生你想要的结果。

数据

我使用了下表作为示例：

CREATE TABLE t AS (
  SELECT DATE '2015-01-01' AS d, 'Same' AS allocation FROM dual UNION ALL
  SELECT DATE '2015-01-02' AS d, 'Good' AS allocation FROM dual UNION ALL
  SELECT DATE '2015-01-03' AS d, 'Same' AS allocation FROM dual UNION ALL
  SELECT DATE '2015-01-04' AS d, 'Same' AS allocation FROM dual UNION ALL  
  SELECT DATE '2015-01-05' AS d, 'Same' AS allocation FROM dual UNION ALL
  SELECT DATE '2015-01-06' AS d, 'Good' AS allocation FROM dual
);

Answer 2

考虑以下查询来解决您的问题：

SELECT COUNT(*) AS `count` FROM test t
WHERE `date` < '2015-01-05' AND allocation = 'Same';

假设给定日期是“2015-01-05”。这里的想法是选择所有小于“2015-01-05”的日期，这意味着它的前几天。由于分配必须是“相同的”，因此它也包含在语句的条件部分中。

Answer 3

试试这个：

SELECT  count(Allocation) as total_allocation FROM table_name 
WHERE (Date BETWEEN CURDATE() - INTERVAL 1 DAY AND CURDATE()) 
AND (Allocation='Same');

这将获取今天日期之前的所有记录，并且其 Allocation='Same'

Answer 4

这更容易：

DEMO

SELECT count(*) AS c
FROM t
WHERE d <
TO_DATE (
       '2015-01-05'
        ,'yyyy-mm-dd'
)
AND allocation = 'Same';