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，表示每天处理多少不同的退款，抛弃任何NULL。

我需要编写哪些SQL才能生成这些结果？

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”，因为您可能只需要计算一次您重复的refundId值，请使用count（DISTINCT refundId）

Answer 3

您使用的是哪种数据库供应商？无论它是什么，用以下相应的构造替换下面的“DateOnly（LogDate）”以从logdate列值中提取日期部分（剥离时间），然后尝试：

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

在Sql server中，例如，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”特定于您未指定的SQL数据库。

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，这使得以下事情成为可能：

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

Answer 6

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

编辑：

如果您不希望退款

，请删除RefundID

Answer 7

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])

SQL group by day，with count

7 个答案: