Python MySQL-按两列分组

我有一张像这样的桌子

id, user_id   , type, date
3,  ivnWvOQqoN, iOS , 2015-11-24 15:46:09
4,  dskIbJhuSd, iOS , 2015-11-23 19:39:31
5,  dskIbJhuSd, iOS , 2015-11-24 23:37:45
6,  ----------, iOS , 2015-11-22 23:38:05
7,  ----------, iOS , 2015-11-23 23:38:10

SELECT COUNT(*) AS entries, user_id, DATE(created_at) as date, device_type
FROM App_Usage
WHERE device_type = 'iOS'
  AND created_at between 2015-11-22 AND 2015-11-25
GROUP BY DATE(created_at), user_id

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 24), 'ios'),
(1L, 'ivnWvOQqoN', datetime.date(2015, 11, 24), 'ios')

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'combo of users' datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(2L, 'combo of users', datetime.date(2015, 11, 24), 'ios'),

现在我正在做这样一个查询

id, user_id   , type, date
3,  ivnWvOQqoN, iOS , 2015-11-24 15:46:09
4,  dskIbJhuSd, iOS , 2015-11-23 19:39:31
5,  dskIbJhuSd, iOS , 2015-11-24 23:37:45
6,  ----------, iOS , 2015-11-22 23:38:05
7,  ----------, iOS , 2015-11-23 23:38:10

SELECT COUNT(*) AS entries, user_id, DATE(created_at) as date, device_type
FROM App_Usage
WHERE device_type = 'iOS'
  AND created_at between 2015-11-22 AND 2015-11-25
GROUP BY DATE(created_at), user_id

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 24), 'ios'),
(1L, 'ivnWvOQqoN', datetime.date(2015, 11, 24), 'ios')

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'combo of users' datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(2L, 'combo of users', datetime.date(2015, 11, 24), 'ios'),

这个返回结果是我所期望的，它看起来是这样的

id, user_id   , type, date
3,  ivnWvOQqoN, iOS , 2015-11-24 15:46:09
4,  dskIbJhuSd, iOS , 2015-11-23 19:39:31
5,  dskIbJhuSd, iOS , 2015-11-24 23:37:45
6,  ----------, iOS , 2015-11-22 23:38:05
7,  ----------, iOS , 2015-11-23 23:38:10

SELECT COUNT(*) AS entries, user_id, DATE(created_at) as date, device_type
FROM App_Usage
WHERE device_type = 'iOS'
  AND created_at between 2015-11-22 AND 2015-11-25
GROUP BY DATE(created_at), user_id

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 24), 'ios'),
(1L, 'ivnWvOQqoN', datetime.date(2015, 11, 24), 'ios')

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'combo of users' datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(2L, 'combo of users', datetime.date(2015, 11, 24), 'ios'),

上面代码中的最后两行将成为下一个代码中的最后一行。上面代码中的第二行变成下面代码中的第二行

我的问题是，我是否可以按日期分组，如果用户不是'----'，那么它将返回如下结果

id, user_id   , type, date
3,  ivnWvOQqoN, iOS , 2015-11-24 15:46:09
4,  dskIbJhuSd, iOS , 2015-11-23 19:39:31
5,  dskIbJhuSd, iOS , 2015-11-24 23:37:45
6,  ----------, iOS , 2015-11-22 23:38:05
7,  ----------, iOS , 2015-11-23 23:38:10

SELECT COUNT(*) AS entries, user_id, DATE(created_at) as date, device_type
FROM App_Usage
WHERE device_type = 'iOS'
  AND created_at between 2015-11-22 AND 2015-11-25
GROUP BY DATE(created_at), user_id

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(1L, 'dskIbJhuSd', datetime.date(2015, 11, 24), 'ios'),
(1L, 'ivnWvOQqoN', datetime.date(2015, 11, 24), 'ios')

((1L, '----------', datetime.date(2015, 11, 22), 'ios'),
(1L, 'combo of users' datetime.date(2015, 11, 23), 'ios'),
(1L, '----------', datetime.date(2015, 11, 23), 'ios'),
(2L, 'combo of users', datetime.date(2015, 11, 24), 'ios'),

我在第一个结果中留下了user_id，以表明只要用户是“----”，我就希望对这些用户进行分组，对其他用户进行分组

所以我想按日期分组，然后按用户不是“-----”的位置分组，我怎么能做到这一点，或者这是可能的

期望输出

输出：

number of uses, user_id**, type, date;
(1, '----------', 'ios', '2015-11-22 23:38:05'),
(1, 'some users or combo', 'ios', '2015-11-22 13:33:33'),
(2, 'some users or combo', 'ios', '2015-11-23 13:35:37'),
(1, '----------', 'ios', '2015-11-24 00:09:44'),
(2, 'some users or combo', 'ios', '2015-11-24 00:09:44'),

**所以我希望任何'----'的用户都是 以“----”分组，但不使用“----”的任何用户 所有其他用户都被占满了。正如你在24日所看到的那样 2个不同的用途被合并为一个，其用途为“----” 这是上面的最后两行

select count(*) as entries, dateCreate, device_type, user_id
from 
    (SELECT DATE(created_at) as dateCreate, device_type, case when user_id = '----------' then '-----------' else 'combo of users' as user_id  
    FROM App_Usage 
    WHERE device_type = 'iOS' 
    AND created_at between 2015-11-22 
    AND 2015-11-25) as temp_table 
GROUP BY dateCreate, user_id, device_type

---

注意：使用dateCreate而不是date，因为它是一个关键字

能否提供SQLFIDLE和适当的result@Strawberry当然，给我一分钟。一般GROUP BY规则说：如果指定了GROUP BY子句，“选择列表中的每个列引用必须标识一个分组列或是一个集合函数的参数。@在使用FIDLE时，我对结果添加了一个编辑以使其更清晰，我希望对“----”用户进行分组，然后对所有其他用户进行分组，如果你看到我说的话mean@jarlh在本例中，这不重要给出错误1248-每个派生表都必须有自己的别名更改要添加的别名？此查询从结果中删除user_id='----'的任何值是否正确？太好了。我在问题的最后一个代码部分对结果添加了一个编辑，以使其更清楚，我希望对“----”用户进行分组，然后对所有其他用户进行分组，如果你明白我的意思，请在编辑中查看所需的输出，此小提琴显示了表中的内容