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我很难理解这一点。我正在使用 MS SQL 2008，并且我有一个名为 Activity 的表，其中包含 3 个字段：customer（客户的身份）、visitDate（他们访问的日期）和 customerType（他们是什么类型的客户）。这是 3 天的数据：

customer    visitDate                   customerType

customer1   2013-10-01 07:00:00.000     A
customer1   2013-10-01 09:00:00.000     A
customer2   2013-10-01 10:00:00.000     B

customer1   2013-10-02 09:00:00.000     A
customer2   2013-10-02 09:00:00.000     B
customer3   2013-10-02 09:00:00.000     B
customer1   2013-10-02 09:00:00.000     A

customer1   2013-10-03 07:00:00.000     A

我想要实现的是编写一个查询，该查询对显示每天的数据进行分组，该查询还计算每天的用户类型，以便结果如下所示：

visitDate   TypeA   TypeB   Total

2013-10-01  1       1       2
2013-10-02  1       2       3
2013-10-03  1       0       0

请注意，如果某人在同一天访问了不止一次，则仅将其计为当天的一次访问。

我知道这与分组有关，但我不知道从哪里开始。

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3 回答 3

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稍微棘手的一点是在给定的日期和类型中只计算一次客户，即使他们当天有多个记录：

select 
   visitDate, 
   sum(case when customerType = 'A' then 1 else 0 end) as TypeA,
   sum(case when customerType = 'B' then 1 else 0 end) as TypeB,
   count(*) as Total
from (
    select distinct 
        customer, 
        cast(visitdate as date) as visitdate, 
        customertype from activity
    ) x
group by 
    visitdate

Example SQLFiddle

于 2013-10-27T22:15:46.983 回答
0 
select visitDate, 
       sum(case when customerType = 'A' then 1 else 0 end) as TypeA,
       sum(case when customerType = 'B' then 1 else 0 end) as TypeB,
       count(*) as Total
from activity
group by visitDate
于 2013-10-27T21:59:47.200 回答
0

这也将减少该DateTime领域的时间部分：

select 
  cast(visitDate as Date) as visitDate, 
  sum(case customerType when 'A' then 1 else 0 end) as TypeA,
  sum(case customerType when 'B' then 1 else 0 end) as TypeB,
  count(*) as Total
from activity
group by cast(visitDate as Date)
于 2013-10-27T22:07:34.860 回答