开发者

What is the accepted way to replace java.util.Date(year,month,day)

I'm trying to do something really simple, but starting to realize that dates in Java are a bit of minefield. All I want is to get passed groups of three ints ( a year, a month and a date) create some Date objects, do some simple test on them (along the lines of as date A before date B and after January 1 1990), convert them to java.sql.Date objects and pass them off to the database via JDBC.

All very simple and works fine using the java.util.Date(int year,int month,int day)开发者_如何学运维 constructor. Of course that constructor is depreciated, and I'd like to avoid using depreciated calls in new code I'm writing. However all the other options to solve this simple problem seem stupidly complicated. Is there really no simple way to do what I want without using depreciated constructors?

I know the standard answer to all Java date related questions is "use joda time", but I really don't want to start pulling in third party libraries for such a seemingly trivial problem.


The idea is to use the Calendar class, like so:

Calendar cal = Calendar.getInstance();
cal.set(year, month, date);
Date date = cal.getTime();

Indeed, if you check the Javadoc of the constructor you are mentioning, it is exactly what is suggested:

Date(int year, int month, int date)
          Deprecated. As of JDK version 1.1, replaced by Calendar.set(year + 1900, month, date) or GregorianCalendar(year + 1900, month, date).

Or ... use JodaTime :-).


Well, you can use the deprecated constructor. It has been deprecated for over 13 years and still works - it isn't going anywhere. If you don't want to do that and don't want to use a third party library, you have to use Calendar.

In Java 7, hopefully there will be a new time API, and you won't have the need for a third party API anymore.


LG's answer is valid but in general you should use Calendar for all your date operations and leave Date out of it unless your API requires it explicitly. In that case you can just convert it when passing it to the API.

0

上一篇:

下一篇:

精彩评论

暂无评论...
验证码 换一张
取 消

最新问答

问答排行榜