日期时间上的floor()函数-SQL Server

http://sqlfiddle.com/#!18/639ec/2

CREATE TABLE TEST
(
ID INT,
OrderNo int,
DateReceived datetime
)

INSERT INTO TEST (ID,OrderNo,DateReceived)
VALUES ('1', '3454', '07-20-2018 00:00:00')

查询:

DECLARE @StartDate datetime,
@EndDate datetime,
@FlooredStart datetime ,
@FlooredEnd datetime 

SET @StartDate = '07-20-18'
SET @EndDate = '07-20-18'
SET @FlooredStart  = CAST(FLOOR(CAST(@startDate AS FLOAT)) AS DATETIME)
SET @FlooredEnd = DATEADD(d, 1, CAST(FLOOR(CAST(@endDate AS FLOAT)) AS DATETIME))

SELECT * FROM TEST 
WHERE DateReceived = @FlooredStart and DateReceived < @FlooredEnd

在我的实时版本中,如果接收日期为空,则默认为当天的12:00:00。因此,对于这个示例,如果我在07-20-18上搜索订单,它将不会返回07-20-18上午12:00:00下的订单。

因此,我想在该子句中添加>=

DECLARE @StartDate datetime,
@EndDate datetime,
@FlooredStart datetime ,
@FlooredEnd datetime 

SET @StartDate = '07-20-18'
SET @EndDate = '07-20-18'
SET @FlooredStart  = CAST(FLOOR(CAST(@startDate AS FLOAT)) AS DATETIME)
SET @FlooredEnd = DATEADD(d, 1, CAST(FLOOR(CAST(@endDate AS FLOAT)) AS 
DATETIME))

SELECT * FROM TEST 
-- WHERE DateReceived BETWEEN @StartDate AND @EndDate
WHERE DateReceived >= @FlooredStart and DateReceived < @FlooredEnd

结果:

| ID | OrderNo |         DateReceived |
|----|---------|----------------------|
|  1 |    3454 | 2018-07-20T00:00:00Z |

我只是想知道我的逻辑是否正确?有人能给我解释一下floor()函数在做什么吗?我知道它是在计算每年/月的第一天,但这里需要吗?我在网上查了一下,找不到确切的答案。

谢谢