sql - SQL 按天分组，有计数

Question

我在 SQL Server 中有一个如下所示的日志表：

CREATE TABLE [dbo].[RefundProcessLog](
 [LogId] [bigint] IDENTITY(1,1) NOT NULL,
 [LogDate] [datetime] NOT NULL,
 [LogType] [varchar](10) COLLATE SQL_Latin1_General_CP1_CI_AS NOT NULL,
 [RefundId] [int] NULL,
 [RefundTypeId] [smallint] NULL,
 [LogMessage] [varchar](1000) COLLATE SQL_Latin1_General_CP1_CI_AS NOT NULL,
 [LoggedBy] [varchar](50) COLLATE SQL_Latin1_General_CP1_CI_AS NOT NULL,
 CONSTRAINT [PK_RefundProcessLog] PRIMARY KEY CLUSTERED 
(
 [LogId] ASC
) ON [PRIMARY]
) ON [PRIMARY]

GO

我想要的是一个结果列表，它代表每天处理了多少不同的退款，丢弃任何 NULL。

我需要编写什么 SQL 才能产生这些结果？

Answer 1

我喜欢（MS SQL）中的这种方法：

SELECT 
  Convert(char(8), LogDate, 112),
  count(distinct RefundId)
FROM RefundProcessing
GROUP BY Convert(char(8), LogDate, 112)

Answer 2

select cast(LogDate as date) as LogDate, count(refundId) as refundCount
from yourTable
group by cast(LogDate as date)

根据您使用的 SQL 方言，您可能需要将 CAST 更改为其他内容。该表达式应将 LogDate 转换为仅日期值。

另外，如果您说“不同的refundId”，因为您可能只想计算一次refundId的重复值，请使用count(DISTINCTrefundId)

Answer 3

您使用的是哪个数据库供应商？不管是哪一种，用适当的结构替换下面的“DateOnly(LogDate)”，以从 logdate 列值中提取日期部分（去掉时间），然后试试这个：

Select [DateOnly(LogDate)], Count Distinct RefundId
From RefundProcessLog
Group By [DateOnly(LogDate)]

例如，在 Sql server 中，适当的构造将是：

Select DateAdd(day, 0, DateDiff(day, 0, LogDate)), Count(Distinct RefundId)
From RefundProcessLog
Group By DateAdd(day, 0, DateDiff(day, 0, LogDate))

Answer 4

SELECT COUNT(RefundId), DateOnly(LogDate) LoggingDate
FROM RefundProcessLog
GROUP BY DateOnly(LogDate)

“DateOnly”特定于您的 SQL 数据库，您没有指定它。

对于 SQL Server，您可以将 DateAdd(dd,0, DateDiff(dd,0,LogDate)) 用于“DateOnly”

Answer 5

SQL Server 2008 引入了date数据类型，使以下操作成为可能：

select convert(date, LogDate),
      ,count(refundid) AS 'refunds'
  from RefundProcessing
group by convert(date,LogDate)
order by convert(date,LogDate)

Answer 6

在 SqlServer 中，它将类似于：

select datepart(YEAR, [LogDate]), datepart(MONTH, [LogDate]), datepart(DAY, [LogDate]), count(refundid) as [Count]
from [RefundProcessing]
group by datepart(YEAR, [LogDate]), datepart(MONTH, [LogDate]), datepart(DAY, [LogDate])

Answer 7

Select count(*), LogDate, refundid from RefundProcessLog
where refundid is not null
group by LogDate, refundid

编辑：

如果您不希望它被退款分解，或者删除 RefundID