Question

我正在努力解决这个问题，这里是表格设置：

date       | time | count
----------------------------
12/12/2015 | 0:00 | 8
12/12/2015 | 1:00 | 19
12/12/2015 | 2:00 | 36
12/13/2015 | 0:00 | 12
12/13/2015 | 1:00 | 22
12/13/2015 | 2:00 | 30
12/14/2015 | 0:00 | 14
12/14/2015 | 1:00 | 26
12/14/2015 | 2:00 | 38

我希望我的查询返回的内容是这样的：

date       | time | count | DAY       | AVG/HR | AVG/DAY 
---------------------------------------------------------
12/12/2015 | 0:00 | 8     | MONDAY    | 11.33  | 63
12/12/2015 | 1:00 | 19    | MONDAY    | 22.33  | 63
12/12/2015 | 2:00 | 36    | MONDAY    | 34.67  | 63
12/13/2015 | 0:00 | 12    | TUESDAY   | 11.33  | 64
12/13/2015 | 1:00 | 22    | TUESDAY   | 22.33  | 64
12/13/2015 | 2:00 | 30    | TUESDAY   | 34.67  | 64
12/14/2015 | 0:00 | 14    | WEDNESDAY | 11.33  | 78
12/14/2015 | 1:00 | 26    | WEDNESDAY | 22.33  | 78
12/14/2015 | 2:00 | 38    | WEDNESDAY | 34.67  | 78

所以基本上就是返回所有行（表中会有数月的数据，每天有24条记录/小时）。并添加星期几字段，每小时计数的平均值以及每周平均每天的计数。最后两个是我正在努力的。以下是我到目前为止的情况：

SELECT DATE, TIME, COUNT,
     TO_CHAR(DATE, 'DAY'),
     (SELECT AVG(t2.COUNT) 
        FROM tableXX t2
       WHERE t2.time = t1.time
    GROUP BY t2.time) AS AvgPerHr
     (SELECT AVG(t2.COUNT) 
        FROM tableXX t2
       WHERE TO_CHAR(t2.DATE, 'DAY') = TO_CHAR(t1.DATE, 'DAY')
    GROUP BY TO_CHAR(t2.DATE, 'DAY')) AS AvgPerDay
FROM tableXX t1
ORDER BY DATE, TO_DATE(TIME, 'hh24:mi') ASC;

任何建议都会受到赞赏，上面的查询会返回数据，但绝对不准确。

Answer 1

这可以通过使用分析函数来解决。

SELECT DATE, TIME, COUNT,
    TO_CHAR(DATE, 'DAY'),
    AVG(t1.COUNT) 
        OVER (PARTITION BY TIME) AS AvgPerHr,
    AVG(t1.COUNT) 
        OVER (PARTITION BY TO_CHAR(DATE, 'DAY'))  AS AvgPerDay
FROM tableXX t1
ORDER BY DATE, TO_DATE(TIME, 'hh24:mi') ASC;

Answer 2

尝试：

SELECT "DATE", "TIME", "COUNT", TO_CHAR(DATE, 'DAY') "DAY,
       avg( "COUNT" ) Over (partition by  "TIME" ) "AVG/HR",
       SUM( "COUNT" ) Over (partition by  "DATE" )  "AVG/DAY"
FROM tablexx
ORDER BY 1;

我使用SUM( "COUNT" )代替AVG( "COUNT" )，因为您示例的第一行中的63似乎是每天的总和，而不是平均值。

使用内部选择和分组依据查询

2 个答案: