Sql 获取值大于平均借款利率的日期

我有一个名为

BOOK

的表（memberId、ISBN、dateBorrowed） 例如：

isbn | memberId | borrowed -------+---------------+-------------+---- 9998-01-101-9 | | 9998-01-101-9 | | 9998-01-101-9 | | 9998-01-101-9 | 1000 | 2018-10-02 9998-01-101-9 | 1010 | 2018-09-04 9998-01-101-9 | 1021 | 2018-09-14 9998-01-101-9 | | 9998-01-101-9 | 1001 | 2018-10-02 我写的另一个查询是计算平均值：

SELECT SUM(dates)/COUNT(borrowed) AS average 
  FROM (
    SELECT borrowed, COUNT(*) AS dates 
      FROM BOOKS
      WHERE borrowed IS NOT NULL GROUP BY borrowed
  ) AS average;

---

现在，如何将这两个续集连接成一个清晰的续集？

这看起来有点像硬件，所以窗口函数可能是不受限制的

SELECT * 
  FROM (
    SELECT BOOK.*,
           CAST(
           COUNT(1) OVER 
             ( PARTITION BY borrowed
             ) AS FLOAT) cntThatDay,
           CAST(
           SUM(1) OVER() AS FLOAT)/ CAST(
               (SELECT COUNT(DISTINCT borrowed)
                  FROM BOOKS 
               ) AS FLOAT) AS totalAverage
      FROM BOOK 
     WHERE borrowed IS NOT NULL
    ) TMP
WHERE cntThatDay >= totalAverage;

---

使用窗口功能可以帮助您：

我的测试数据：

isbn             borrowed
9998-01-101-1    2018-08-01
9998-01-101-2    2018-08-01
9998-01-101-3    2018-08-01
9998-01-101-4    2018-08-01
9998-01-101-5    2018-08-01
9998-01-101-1    2018-08-02
9998-01-101-2    2018-08-02
9998-01-101-3    2018-08-02
9998-01-101-4    2018-08-03
9998-01-101-5    2018-08-03
9998-01-101-1    2018-08-04
9998-01-101-2    2018-08-04
9998-01-101-3    2018-08-04
9998-01-101-4    2018-08-04
9998-01-101-5    2018-08-05
9998-01-101-1    2018-08-05

   SELECT 
        * 
    FROM (
        SELECT 
            *, 
            borrowed_all_time::decimal / COUNT(*) OVER () as avg_borrows_per_day    -- D
        FROM (
            SELECT DISTINCT                                                         -- C
                borrowed,
                COUNT(*) OVER (PARTITION BY borrowed) as borrowed_on_day,           -- A
                COUNT(*) OVER () as borrowed_all_time                               -- B
            FROM book
        )s 
    )s
    WHERE borrowed_on_day > avg_borrows_per_day                                     -- E

borrowed
2018-08-01
2018-08-04

查询：

isbn             borrowed
9998-01-101-1    2018-08-01
9998-01-101-2    2018-08-01
9998-01-101-3    2018-08-01
9998-01-101-4    2018-08-01
9998-01-101-5    2018-08-01
9998-01-101-1    2018-08-02
9998-01-101-2    2018-08-02
9998-01-101-3    2018-08-02
9998-01-101-4    2018-08-03
9998-01-101-5    2018-08-03
9998-01-101-1    2018-08-04
9998-01-101-2    2018-08-04
9998-01-101-3    2018-08-04
9998-01-101-4    2018-08-04
9998-01-101-5    2018-08-05
9998-01-101-1    2018-08-05

   SELECT 
        * 
    FROM (
        SELECT 
            *, 
            borrowed_all_time::decimal / COUNT(*) OVER () as avg_borrows_per_day    -- D
        FROM (
            SELECT DISTINCT                                                         -- C
                borrowed,
                COUNT(*) OVER (PARTITION BY borrowed) as borrowed_on_day,           -- A
                COUNT(*) OVER () as borrowed_all_time                               -- B
            FROM book
        )s 
    )s
    WHERE borrowed_on_day > avg_borrows_per_day                                     -- E

borrowed
2018-08-01
2018-08-04

答：此窗口函数统计每个

借用的
日期的行数

B:这个窗口函数统计所有行，等于统计所有时间的借用

目前的结果如下：

borrowed      borrowed_on_day   borrowed_all_time
2018-08-01    5                 16
2018-08-01    5                 16
2018-08-01    5                 16
2018-08-01    5                 16
2018-08-01    5                 16
2018-08-02    3                 16
2018-08-02    3                 16
2018-08-02    3                 16
2018-08-03    2                 16
2018-08-03    2                 16
2018-08-04    4                 16
2018-08-04    4                 16
2018-08-04    4                 16
2018-08-04    4                 16
2018-08-05    2                 16
2018-08-05    2                 16

C:因为我们不需要复制品，所以我们用
不同的

D：在消除所有绑定行之后计算所有行将给出不同天数的计数。这除以所有时间的借款得出平均每天的借款。
decimal
cast是必需的。它将整数除法（
16/5==3
）转换为浮点除法（
16/5==3.2
）

E:现在我们可以过滤每天的借款>每天平均借款

结果：

isbn             borrowed
9998-01-101-1    2018-08-01
9998-01-101-2    2018-08-01
9998-01-101-3    2018-08-01
9998-01-101-4    2018-08-01
9998-01-101-5    2018-08-01
9998-01-101-1    2018-08-02
9998-01-101-2    2018-08-02
9998-01-101-3    2018-08-02
9998-01-101-4    2018-08-03
9998-01-101-5    2018-08-03
9998-01-101-1    2018-08-04
9998-01-101-2    2018-08-04
9998-01-101-3    2018-08-04
9998-01-101-4    2018-08-04
9998-01-101-5    2018-08-05
9998-01-101-1    2018-08-05

   SELECT 
        * 
    FROM (
        SELECT 
            *, 
            borrowed_all_time::decimal / COUNT(*) OVER () as avg_borrows_per_day    -- D
        FROM (
            SELECT DISTINCT                                                         -- C
                borrowed,
                COUNT(*) OVER (PARTITION BY borrowed) as borrowed_on_day,           -- A
                COUNT(*) OVER () as borrowed_all_time                               -- B
            FROM book
        )s 
    )s
    WHERE borrowed_on_day > avg_borrows_per_day                                     -- E

borrowed
2018-08-01
2018-08-04

欢迎来到SO。我们是来帮助你的，不是为你工作。通过编写一个sql查询：）到目前为止，您做了哪些尝试？出了什么问题？我已经选择了日期以及它被选择了多少次：选择借来的，计数（*）作为借来的不为空的书中的日期按借来分组；更新了我所做的描述。