Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
297 views
in Technique[技术] by (71.8m points)

sql server - How do I calculate the last day of the month in SQL?

Specifically MSSQL 2005.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Here's a solution that gives you the last second of the current month. You can extract the date part or modify it to return just the day. I tested this on SQL Server 2005.

select dateadd( s, -1, dateadd( mm, datediff( m, 0, getdate() ) + 1, 0 ) );

To understand how it works we have to look at the dateadd() and datediff() functions.

DATEADD(datepart, number, date)  
DATEDIFF(datepart, startdate, enddate)

If you run just the most inner call to datediff(), you get the current month number since timestamp 0.

select datediff(m, 0, getdate() );  
1327

The next part adds that number of months plus 1 to the 0 timestamp, giving you the starting point of the next calendar month.

select dateadd( mm, datediff( m, 0, getdate() ) + 1, 0 );
2010-09-01 00:00:00.000

Finally, the outer dateadd() just subtracts one second from the beginning timestamp of next month, giving you the last second of the current month.

select dateadd( s, -1, dateadd( mm, datediff( m, 0, getdate() ) + 1, 0 ) );
2010-08-31 23:59:59.000


This old answer (below) has a bug where it doesn't work on the last day of a month that has more days than the next month. I'm leaving it here as a warning to others.

Add one month to the current date, and then subtract the value returned by the DAY function applied to the current date using the functions DAY and DATEADD.

dateadd(day, -day(getdate()), dateadd(month, 1, getdate()))

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...