重新实现cume_dist（）

据我所知，用现有的cume_dist()函数无法做到这一点，但不用cume_dist()实现相同的functionalify是相当容易的。
postgres docs表示cume_dist()
返回累积分布，即（当前行之前的分区行数或与当前行对等的分区行数）/（分区总行数）。因此，该值的范围从1/N到1。
下面是我如何设置我的演示数据：

CREATE TABLE my_table (
  target VARCHAR,
  question VARCHAR,
  value INT,
  current BOOL
);

INSERT INTO my_table VALUES
('alice',   'number of hats', 7, true),
('alice',   'number of hats', 6, false),
('alice',   'number of hats', 3, false),
('bob',     'number of hats', 4, true),
('charles', 'number of hats', 9, true),
('david',   'number of hats', 0, true),
('erin',    'number of hats', 4, true),
('alice',   'number of fingers', 10, true),
('bob',     'number of fingers', 10, true),
('charles', 'number of fingers',  9, true),
('charles', 'number of fingers', 10, false),
('david',   'number of fingers', 10, true),
('erin',    'number of fingers', 10, true);

下面是如何在不使用cume_dist()的情况下执行相同的操作（前面的行）/（总行数），并添加一些逻辑来排除不是当前的行：

SELECT
  target,
  question,
  value,
  current,
  (
    -- this is the count of partition rows preceding this row
    SELECT COUNT(1)
    FROM my_table AS inner_table
    WHERE
      inner_table.question = outer_table.question AND
      current = true AND
      inner_table.value <= outer_table.value
  )::real / (
    -- this is the total number of rows for this question
    SELECT COUNT(1)
    FROM my_table AS inner_table
    WHERE
      inner_table.question = outer_table.question AND
      current = true
  )::real * 100.0 AS percentile
FROM my_table AS outer_table;

Here是此查询的一个小链接