使用CASE检查是否应该使用小时零件。
查询
DECLARE @Hours AS INT
DECLARE @SubtractDate AS DATETIME
SET @Hours=24/* User input to have hours */
SET @SubtractDate=DATEADD(hh,@Hours,GETDATE()) - GETDATE()
if @Hours>=24
SELECT
CONVERT(VARCHAR(10),DATEDIFF(DAY,'1900-01-01',@SubtractDate))+ ' Day(s) ' +
+
CASE
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) > 0
THEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) + ' Hour(s)'
ELSE ''
END
AS [Result]
Else
SELECT
CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate))+ ' Hour(s) '
AS [Result]
结果集@时间= 24
| Result |
|-----------|
| 1 Day(s) |
结果集@小时= 25
| Result |
|--------------------|
| 1 Day(s) 1 Hour(s) |
结果SET @小时= 48
| Result |
|-----------|
| 2 Day(s) |
编辑:输出格式化
查询
DECLARE @Hours AS INT
DECLARE @SubtractDate AS DATETIME
SET @Hours=1/* User input to have hours */
SET @SubtractDate=DATEADD(hh,@Hours,GETDATE()) - GETDATE()
if @Hours>=24
SELECT
CASE
WHEN CONVERT(VARCHAR(10),DATEDIFF(DAY,'1900-01-01',@SubtractDate)) = 1
THEN CONVERT(VARCHAR(10),DATEDIFF(DAY,'1900-01-01',@SubtractDate))+ ' Day '
WHEN CONVERT(VARCHAR(10),DATEDIFF(DAY,'1900-01-01',@SubtractDate)) >= 2
THEN CONVERT(VARCHAR(10),DATEDIFF(DAY,'1900-01-01',@SubtractDate))+ ' Days '
END
+
CASE
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) > 0
THEN CASE
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) = 1
THEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) + ' Hour'
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) >= 2
THEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) + ' Hours'
END
ELSE ''
END
AS [Result]
Else
SELECT
CASE
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) > 0
THEN CASE
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) = 1
THEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) + ' Hour'
WHEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) >= 2
THEN CONVERT(VARCHAR(10),DATEPART(hh,@SubtractDate)) + ' Hours'
END
ELSE ''
END
AS [Result]
结果集@小时= 1
| Result |
|--------|
| 1 Hour |
结果集@小时= 2
| Result |
|---------|
| 2 Hours |
结果集@时间= 24
| Result |
|--------|
| 1 Day |
结果集@小时= 25
| Result |
|--------------------|
| 1 Day 1 Hour |
结果集@小时数= 26
| Result |
|---------------|
| 1 Day 2 Hours |
结果集@小时= 48
| Result |
|-----------|
| 2 Days |
(1)编辑你的问题,并提供样本数据,预期的效果,和你想要什么解释。 (2)确定你正在使用的数据库。该代码不是MySQl代码。 –
....这段代码最可能是SQL-server SQL。SQL-server支持函数DATEADD,DATEGET,DATEDIFF和DATEPART,并按照你的方式和用法声明用户变量。 –
当输入:SET @ Hours = 24/*用户输入有小时* /期望的结果将是:1天,而不是现在它显示:1天0小时........ ..............................另外:输入时:SET @ Hours = 48/*用户输入有小时* /期望结果将是:2天,而现在它显示:2天0小时 – neer0097