Sql 按日期和条目计数分组

我要简短一点，这张桌子是这样的：

| id (int) | registerDate (DATETIME)
|----------|-----------------
| 1        | 2014-07-29 12:00:00
| 2        | 2014-08-01 12:00:00
| 3        | 2014-08-01 12:00:00
| 4        | 2014-08-01 12:00:00
| 5        | 2014-08-02 12:00:00
| 6        | 2014-08-02 12:00:00
| 7        | 2014-08-04 12:00:00

| registerDate (DATETIME) | count (int)
| 2014-08-04              | 1
| 2014-08-03              | 0
| 2014-08-02              | 2
| 2014-08-01              | 1
| 2014-07-31              | 0
| 2014-07-30              | 0
| 2014-07-29              | 1

select registerDate  = convert(date,t.registerDate) ,
       registrations = count(*)
from dbo.my_special_registration_table t
where t.registrationDate >= dateadd(day,-7,convert(date,getdate()))
group by convert(date,t.registerDate)
order by 1

如果今天是2014-08-05，我想要这样的结果：

| id (int) | registerDate (DATETIME)
|----------|-----------------
| 1        | 2014-07-29 12:00:00
| 2        | 2014-08-01 12:00:00
| 3        | 2014-08-01 12:00:00
| 4        | 2014-08-01 12:00:00
| 5        | 2014-08-02 12:00:00
| 6        | 2014-08-02 12:00:00
| 7        | 2014-08-04 12:00:00

| registerDate (DATETIME) | count (int)
| 2014-08-04              | 1
| 2014-08-03              | 0
| 2014-08-02              | 2
| 2014-08-01              | 1
| 2014-07-31              | 0
| 2014-07-30              | 0
| 2014-07-29              | 1

select registerDate  = convert(date,t.registerDate) ,
       registrations = count(*)
from dbo.my_special_registration_table t
where t.registrationDate >= dateadd(day,-7,convert(date,getdate()))
group by convert(date,t.registerDate)
order by 1

所以我想知道过去一周（每天）的注册用户数。 我试着在谷歌上找到它（没有成功）-但是，我希望你能帮助我

SELECT registerDate, count(registerDate) FROM [TABLE] WHERE
registerDate between (GETDATE()-7) and GETDATE()   
group by registerDate  
order by registerDate desc

这将采用如下表格：

2   |1905-06-26 00:00:00.000
4   |2014-08-03 00:00:00.000
5   |2014-08-02 00:00:00.000
1   |2014-08-01 00:00:00.000
3   |2014-07-01 00:00:00.000
6   |2010-07-01 00:00:00.000
7   |2015-07-01 00:00:00.000
8   |2014-08-28 00:00:00.000
9   |2014-08-26 00:00:00.000
10  |2014-08-26 00:00:00.000

并创建：

2014-08-28 00:00:00.000 |   1
2014-08-26 00:00:00.000 |   2

问题是它没有显示表中没有的日期。 再给我一点时间，我会有一个更新的版本

编辑：

现在更复杂的一个

-- Declare how far back you want to go
DECLARE @DAYSBACK int = 6

-- Select into a temptable
select CONVERT(date, registerDate) as RegDate, count(registerDate) as DateCount 
  INTO #temptable 
  from Temp where registerDate between (GETDATE()-6) and GETDATE()
  group by registerDate order by registerDate desc

-- Check to see if exists if not, insert row
WHILE @DAYSBACK >= 0 BEGIN
    IF NOT EXISTS (select top 1 1 from #temptable 
                   where RegDate= CONVERT(date, (GETDATE()-@DAYSBACK))
                   group by RegDate)
        INSERT INTO #temptable values ((GETDATE()-@DAYSBACK), 0)
    SET @DAYSBACK = @DAYSBACK -1
END

-- Select what you want
select * from #temptable order by RegDate desc

-- Drop the table you created.
DROP TABLE #temptable

使用与上面相同的表，它将输出：

Register Date | Date Count
--------------------------
2014-08-28    | 1
2014-08-27    | 0
2014-08-26    | 2
2014-08-25    | 0
2014-08-24    | 0
2014-08-23    | 0
2014-08-22    | 0

---

试着这样做：

| id (int) | registerDate (DATETIME)
|----------|-----------------
| 1        | 2014-07-29 12:00:00
| 2        | 2014-08-01 12:00:00
| 3        | 2014-08-01 12:00:00
| 4        | 2014-08-01 12:00:00
| 5        | 2014-08-02 12:00:00
| 6        | 2014-08-02 12:00:00
| 7        | 2014-08-04 12:00:00

| registerDate (DATETIME) | count (int)
| 2014-08-04              | 1
| 2014-08-03              | 0
| 2014-08-02              | 2
| 2014-08-01              | 1
| 2014-07-31              | 0
| 2014-07-30              | 0
| 2014-07-29              | 1

select registerDate  = convert(date,t.registerDate) ,
       registrations = count(*)
from dbo.my_special_registration_table t
where t.registrationDate >= dateadd(day,-7,convert(date,getdate()))
group by convert(date,t.registerDate)
order by 1

如果您尝试使用类似于

datediff（）

的方法过滤7天以上的注册：

---

datediff（day，t.registrationDate，getdate（））在哪里可以说清楚？2014-08-01 12:00:00出现了三次，但在您的结果中，它显示1是数据库中的真实表（这会创建表吗）？然而，结果显然符合我的期望。谢谢你。我是否必须为此创建一个过程（关于MySQL，我使用的是phpmyadmin）。@BlackHat
#tentable
创建一个本地临时表，该表仅在当前会话中可见。我可能会把它放到一个存储过程中，它会为它创建一个计划，并且每次运行时都会更快。一些简短的阅读很好地解释了这一点。关于#,##和@的类似问题：