Home > Database > Mysql Tutorial > body text

oracle查看当前日期是第几个星期的方法

WBOY
Release: 2016-06-07 17:56:02
Original
1571 people have browsed it

oracle查看当前日期是第几个星期方法的代码段,需要的朋友可以参考一下

系统当前时间是多少周,就是是今年的第几个星期
代码如下:
select to_char(sysdate,'ww') ,to_char(sysdate,'iw') from dual;
select to_char(sysdate,'ddd') from dual;
select TRUNC(SYSDATE,'MM') from dual;1)

ww的算法为每年1月1日为第一周开始,date+6为每一周结尾
例如:20050101为第一周的第一天,而第一周的最后一天为20050101+6=20050107  
公式: 每周第一天 :date + 周 * 7 - 7  每周最后一天:date + 周 * 7 - 12)

iw的算法为星期一至星期日算一周,且每年的第一个星期一为第一周,
例如:20050101为星期六,所以用iw的算法是前年的53周,而20050103之后才是第一周的开始。  
公式: 每周第一天 :next_day(date) + 周 * 7 - 7每周最后一天:next_day(date) + 周 * 7 - 13)

其它:  
A、查今天是 "本月" 的第几周  SELECT TO_CHAR(SYSDATE,'WW') - TO_CHAR(TRUNC(SYSDATE,'MM'),'WW') + 1 AS "weekOfMon" from dual;  或  SELECT TO_CHAR(SYSDATE,'W') AS "weekOfMon" from dual;  
B、查今天是 "今年" 的第几周  select to_char(sysdate,'ww') from dual;  或  select to_char(sysdate,'iw') from dual;
统计哪年哪个星期的哪一天:
代码如下:
int year=2011;
int week=1;
Calendar calFirstDayOfTheYear = new GregorianCalendar(year,
Calendar.JANUARY, 1);
calFirstDayOfTheYear.add(Calendar.DATE, 7 * (week-1));


int dayOfWeek = calFirstDayOfTheYear.get(Calendar.DAY_OF_WEEK);


Calendar calFirstDayInWeek = (Calendar)calFirstDayOfTheYear.clone();
calFirstDayInWeek.add(Calendar.DATE,
calFirstDayOfTheYear.getActualMinimum(Calendar.DAY_OF_WEEK) - dayOfWeek);
Date firstDayInWeek = calFirstDayInWeek.getTime();
System.out.println(year + "年第" + week + "个礼拜的第一天是" + ReportDateUtil.getFromatDay().format(firstDayInWeek));


Calendar calLastDayInWeek = (Calendar)calFirstDayOfTheYear.clone();
calLastDayInWeek.add(Calendar.DATE,
calFirstDayOfTheYear.getActualMaximum(Calendar.DAY_OF_WEEK) - dayOfWeek);
Date lastDayInWeek = calLastDayInWeek.getTime();
System.out.println(year + "年第" + week + "个礼拜的最后一天是" + ReportDateUtil.getFromatDay().format(lastDayInWeek));
Related labels:
source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template
About us Disclaimer Sitemap
php.cn:Public welfare online PHP training,Help PHP learners grow quickly!