我有以下日期时间：
Start: 2013-1-1 00:00:00.000
End: 2013-1-31 23:59:59.599
我如何在每个时期的分段中将其分成int段。
例如，如果天数= 1：
2013-1-1 00:00:00.000-> 01-01-2013 23:59:59
2013-1-2 00:00:00.000 -> 2013-1-2 23:59:59
2013-1-3 00:00:00.000 -> 2013-1-3 23:59:59
etc...
我已经为c＃阅读了这篇文章Split period into month parts，我试着这样做：
private static ArrayList<Period> splitPeriods(int days){
ArrayList<Period> list = new ArrayList<>();
java.sql.Timestamp start = new Timestamp(110, 0, 1, 00, 00, 00, 00);
java.sql.Timestamp end = new Timestamp( 110, 2, 1, 00, 00, 00, 00);
Timestamp current = start;
Period period;
while(current.before(end)){
period = new Period();
period.setStart(current);
current.setDate((current.getDate()+days));
period.setEnd(current);
System.out.println(period.getStart() +"\t"+period.getEnd());
}
return list;
}
课程时间：
public class Period {
private java.sql.Timestamp end;
private java.sql.Timestamp start;
//..set - get
}
但是期间的输出是：
2013-1-2 00:00:00.000 -> 2013-1-2 00:00:00.000
2013-1-3 00:00:00.000 -> 2013-1-3 00:00:00.000
etc...
有什么想法吗？
答案 0 :(得分：1)
我找到了解决方案。 Ι为开始时段和结束时段创建新的时间戳。 开始周期采用递增前的当前时间戳的值。 结束时段采用递增后的当前时间戳的值。
代码：
期间等级
public class Period {
private java.sql.Timestamp end;
private java.sql.Timestamp start;
//..set - get
}
主要课程
public class MainClass{
public static void main(String[] args) {
ArrayList<Period> list = splitPeriods(15);
}
private static ArrayList<Period> splitPeriods(int days){
ArrayList<Period> list = new ArrayList<>();
java.sql.Timestamp start = new Timestamp(110, 2, 1, 00, 00, 00, 00); //start 1 March 2010
java.sql.Timestamp end = new Timestamp( 110, 3, 30, 00, 00, 00, 00); //end 30 April 2010
Timestamp current = start;
Period period;
while(current.before(end)){
period = new Period();
//new timestamp for start Period
java.sql.Timestamp startPeriod = new Timestamp(current.getTime());
//change current timestamp
current.setDate((current.getDate()+days));
//new timestamp for end period.
//set prev day and time 23:59:59...
java.sql.Timestamp endPeriod = new Timestamp(current.getTime());
endPeriod.setDate(endPeriod.getDate()-1);
endPeriod.setHours(23);
endPeriod.setMinutes(59);
endPeriod.setSeconds(59);
endPeriod.setNanos(599999999);
//add to list
period.setStart(startPeriod);
period.setEnd(endPeriod);
System.out.println(period.getStart() +"\t"+period.getEnd());
}
return list;
}
}
结果：
input
start: 1 March 2010
end: 30 April 2010
output
2010-03-01 00:00:00.0 2010-03-15 23:59:59.599999999
2010-03-16 00:00:00.0 2010-03-30 23:59:59.599999999
2010-03-31 00:00:00.0 2010-04-14 23:59:59.599999999
2010-04-15 00:00:00.0 2010-04-29 23:59:59.599999999
答案 1 :(得分：0)
这是一个棘手的正则表达式问题！我认为这种困难应被视为正则表达可能不是最佳工具的证据。
激活multiline flag后，此正则表达式将起作用：
Start: ([0-9]{4}(?:-[0-9]{1,2}){2} +(?:[0-9]{2}:){2}[0-9]{2})(?:\.[0-9]{2,3})?\nEnd: ([0-9]{4}(?:-[0-9]{1,2}){2} +(?:[0-9]{2}:){2}[0-9]{2})(?:\.[0-9]{2,3})?\n $1 -> $2 自由间隔：
Start: ([0-9]{4}：Start，（开始捕获组），4位数字(?:[0-9]{2}:){2}[0-9]{2})：两位数字然后冒号（两次），后跟两位数字(?:\.[0-9]{2,3})?：可选：点后跟两位或三位\r\n：Newline End: ...：End后跟与Start 输入：
Start: 2013-1-1 00:00:00.000
End: 2013-1-31 23:59:59.59
Start: 2013-1-1 00:00:00.000
End: 2013-1-1 23:59:59
Start: 2013-1-2 00:00:00.000
End: 2013-1-2 23:59:59
Start: 2013-1-3 00:00:00.000
End: 2013-1-3 23:59:59
爪哇：
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class StartEndDateLinesToArrowSingleLine {
private static final String LINE_SEP = System.getProperty("line.separator", "\r\n");
public static final void main(String[] ignored) {
String regexMain = "([0-9]{4}(?:-[0-9]{1,2}){2} +" +
"(?:[0-9]{2}:){2}[0-9]{2})" +
"(?:\\.[0-9]{2,3})?" + LINE_SEP;
String findWhatRegex = "Start: " + regexMain + "End: " + regexMain;
String rplcWith = "$1 -> $2";
String input = new StringBuilder().
append("Start: 2013-1-1 00:00:00.000").append(LINE_SEP).
append("End: 2013-1-31 23:59:59.599").append(LINE_SEP).
append("Start: 2013-1-1 00:00:00.000").append(LINE_SEP).
append("End: 2013-1-1 23:59:59" ).append(LINE_SEP).
append("Start: 2013-1-2 00:00:00.000").append(LINE_SEP).
append("End: 2013-1-2 23:59:59" ).append(LINE_SEP).
append("Start: 2013-1-3 00:00:00.000").append(LINE_SEP).
append("End: 2013-1-3 23:59:59" ).append(LINE_SEP).toString();
Matcher mtchr = Pattern.compile(findWhatRegex, Pattern.MULTILINE).matcher(input);
StringBuffer bfr = new StringBuffer();
while(mtchr.find()) {
bfr.setLength(0);
mtchr.appendReplacement(bfr, rplcWith);
bfr.append(LINE_SEP);
mtchr.appendTail(bfr);
mtchr.reset(bfr.toString());
}
System.out.println(bfr.toString());
}
}
输出：
2013-1-1 00:00:00 -> 2013-1-31 23:59:59
2013-1-1 00:00:00 -> 2013-1-1 23:59:59
2013-1-2 00:00:00 -> 2013-1-2 23:59:59
2013-1-3 00:00:00 -> 2013-1-3 23:59:59
答案 2 :(得分：0)
private List<DaySlot> getDaySlots(Date start, Date end) {
List<DaySlot> ret = new ArrayList<>();
Date current = start;
while(current.before(end)) {
Date startOfTheDay = getStartOfTheDay(current);
Date endOfTheDay = getEndOfTheDay(startOfTheDay);
ret.add(new DaySlot(current, endOfTheDay.before(end)? endOfTheDay: end));
current = getNextDay(startOfTheDay);
}
return ret;
}
private LocalDate getDateOnly(Date date) {
return date.toInstant().atZone(ZoneId.systemDefault()).toLocalDate();
}
private Date getStartOfTheDay(Date current) {
return Date.from(getDateOnly(current).atStartOfDay(ZoneId.systemDefault()).toInstant());
}
private Date getNextDay(Date startOftheDay) {
Calendar calendar = Calendar.getInstance();
calendar.setTime(startOftheDay);
calendar.add(Calendar.DATE, 1);
System.out.println(calendar.getTime());
return calendar.getTime();
}
private Date getEndOfTheDay(Date startOftheDay) {
Calendar calendar = Calendar.getInstance();
calendar.setTime(startOftheDay);
calendar.add(Calendar.HOUR_OF_DAY, 23);
calendar.add(Calendar.MINUTE, 59);
calendar.add(Calendar.SECOND, 59);
calendar.add(Calendar.MILLISECOND, 599);
System.out.println(calendar.getTime());
return calendar.getTime();
}
电话如下：
List<DaySlot> daySlots = getDaySlots(durationStartTime, durationEndTime);