SQL date issue! Problems with sqlserver date calculations

dateadd(dd, datediff(dd,0,getdate())0) means to return the current date.

dd is the date component representing day, and there are others such as month, year, etc.

dateadd is the time to calculate the number of date components before or after a date, such as the time of 10 days before today.

dateadd(dd,'2012-10-26', -10), and the next 10 days are dateadd(dd,'2012-10-26',10)

datediff calculates the number of date components before two dates, such as the number of days between two dates: datediff(dd,'2012-9-10','2012-9-11')

The 0 here represents'1900-1-1'

The whole expression is equivalent to dateadd(dd, datediff(dd,'1900-1-1',getdate())'1900-1-1')

That is, the number of days between the current system time and 1900-1-1 is calculated first, and then the date of 1900-1-1 plus this number of days is calculated.

, number, date)

Returns a variant (date) with a date plus a time interval.

date1, date2)

Represents the time interval used to calculate the time difference between date1 and date2, datediff(dd,0,getdate())0) from the inside out.

datediff(dd,0,getdate()) is the subtraction of two dates.

dateadd(dd, datediff(dd,0,getdate())0)

1、dateadd(dd, datediff(dd,0,getdate())0)

1) getdate(): A function that returns the current date and time.

2) DateDiff(dd,0,getDate()) is to calculate the number of days between the current date and the date of "19000101. (Period and time variables are calculated from "19000101 as in milliseconds.)

So the datediff function specifies that the first time expression is "0". 3) dateadd, which increases the number of days from the current date to "19000101".

1. The date and time of the current system.

select getdate()

select dateadd(day,2,'2004-10-15'- Return: 2004-10-17

3. DateDiff returns the number of date and time boundaries that span two specified dates.

select datediff(day,'2004-09-01','2004-09-18'- returned: 17

4. datepart returns an integer representing the specified date part of the specified date.

select datepart(month, '2004-10-15'- Return 10

5. datename returns a string that represents the specified date part of the specified date.

select datename(weekday, '2004-10-15'- Return: Friday.

6. day(),month(),year() can be compared with datepart.

select current-date=convert(varchar(10),getdate(),120),select currenttime=convert(varchar(8),getdate(),114),select datename(dw,'2004-10-15')select the number of weeks in the year = datename(week,'2004-10-15'),select Today is the day of the week =datename(weekday,'2004-10-15')

7. Find the difference in days.

select datediff(day,'2004-01-01',getdate())