Java 把小时和分钟加起来
把小时和分钟加在一起会得出错误的答案 这里有两个小时的列表 [[0小时:0分钟:0秒,1小时:16分钟:22秒,0小时:1分钟:53秒,23小时:54分钟:18秒],[0小时:8分钟:22秒,0小时:8分钟:22秒,0小时:8分钟:22秒]]这是小时 然后,我将其划分为子列表,对每个子列表进行计算。这是其中的一个子列表 [0小时:0分钟:0秒,1小时:16分钟:22秒,0小时:1分钟:53秒,23小时:54分钟:18秒] 因此,我得到的这个子列表的答案是:1小时:12分钟:33秒,这是不正确的。我想大约25小时外加几分钟Java 把小时和分钟加起来,java,time,java-time,localtime,datetimeformatter,Java,Time,Java Time,Localtime,Datetimeformatter,把小时和分钟加在一起会得出错误的答案 这里有两个小时的列表 [[0小时:0分钟:0秒,1小时:16分钟:22秒,0小时:1分钟:53秒,23小时:54分钟:18秒],[0小时:8分钟:22秒,0小时:8分钟:22秒,0小时:8分钟:22秒]]这是小时 然后,我将其划分为子列表,对每个子列表进行计算。这是其中的一个子列表 [0小时:0分钟:0秒,1小时:16分钟:22秒,0小时:1分钟:53秒,23小时:54分钟:18秒] 因此,我得到的这个子列表的答案是:1小时:12分钟:33秒,这是不正确的。
public List<String> getTheHoursWorked() {
DateTimeFormatter parser = DateTimeFormatter.ofPattern("H 'hours :' m 'mins :' s 'sec'", Locale.US);
final DateFormat dt = new SimpleDateFormat("HH:mm:ss");
final Calendar c = Calendar.getInstance(TimeZone.getDefault(), Locale.getDefault());
c.clear();
long startingMS = c.getTimeInMillis();
int counter = 0;
for (int k = 0; k < hours.size(); k++){
List<Object> shorter = new ArrayList<>();
List<Object> temp;
temp = (List<Object>) hours.get(k);
long milliseconds = 0;
for (int m = 0; m < shorter.size(); m++) {
LocalTime odt = LocalTime.parse((CharSequence) shorter.get(m), parser);
DateTimeFormatter formatter =
DateTimeFormatter.ofPattern("HH:mm:ss");
String printDate = formatter.format(odt);
try {
milliseconds = milliseconds + (dt.parse(printDate).getTime() - startingMS);
System.out.println(milliseconds + "MILISECONDS");
} catch (ParseException e) {
e.printStackTrace();
}
}
hoursToString.add(String.valueOf(shorter));
String s = milliseconds / 1000 % 60 + " seconds";
String m = milliseconds /(60 * 1000) % 60 + " minutes";
String h = milliseconds / (60 * 60 * 1000) % 24 + " hours";
String together = h+":"+m+":"+s;
togetherHours.add(together);
}
return togetherHours;
}
公共列表getTheHoursWorked(){
DateTimeFormatter parser=DateTimeFormatter.of模式(“H'小时:'m'分钟:'s'秒',Locale.US);
最终日期格式dt=新的简化格式(“HH:mm:ss”);
final Calendar c=Calendar.getInstance(TimeZone.getDefault(),Locale.getDefault());
c、 清除();
长启动ms=c.getTimeInMillis();
int计数器=0;
对于(int k=0;k
public static List<String> getTheHoursWorked() {
final DateTimeFormatter parser = DateTimeFormatter.ofPattern("H 'hours :' m 'mins :' s 'sec'", Locale.US);
final List<String> togetherHours = new LinkedList<>();
for (int k = 0; k < hours.size(); k++){
final List<String> shorter = hours.get(k);
long seconds = 0;
for (int m = 0; m < shorter.size(); m++) {
final LocalTime odt = LocalTime.parse(shorter.get(m), parser);
seconds += odt.getHour() * 3600 + odt.getMinute() * 60 + odt.getSecond();
}
final String s = (seconds ) % 60 + " seconds";
final String m = (seconds / 60) % 60 + " minutes";
final String h = (seconds / 3600) + " hours";
final String together = h+":"+m+":"+s;
togetherHours.add(together);
}
return togetherHours;
}
public static List<String> getTheHoursWorked_new() {
final DateTimeFormatter parser = DateTimeFormatter.ofPattern("H 'hours :' m 'mins :' s 'sec'", Locale.US);
final List<String> togetherHours = new LinkedList<>();
for (final var shorter: hours){
long seconds = 0;
for (final var duration : shorter) {
final LocalTime odt = LocalTime.parse(duration, parser);
seconds += odt.getHour() * 3600 + odt.getMinute() * 60 + odt.getSecond();
}
final String s = (seconds ) % 60 + " seconds";
final String m = (seconds / 60) % 60 + " minutes";
final String h = (seconds / 3600) + " hours";
final String together = h+":"+m+":"+s;
togetherHours.add(together);
}
return togetherHours;
}
公共静态列表getTheHoursWorked(){
final DateTimeFormatter parser=DateTimeFormatter.of模式(“H'小时:'m'分钟:'s'秒',Locale.US);
最终列表togetherHours=新建LinkedList();
对于(int k=0;k
解决方案的步骤应该是
0:0的时间开始
import java.time.LocalTime;
import java.time.format.DateTimeFormatter;
import java.util.List;
public class Main {
public static void main(String[] args) {
List<String> list = List.of("0 hours : 0 mins : 0 sec", "1 hours : 16 mins : 22 sec",
"0 hours : 1 mins : 53 sec", "23 hours : 54 mins : 18 sec");
// Start with a time of 0:0
int sumHours = 0;
int sumMinutes = 0;
int sumSeconds = 0;
// Iterate the list and add all hours, minutes and seconds separately after
// parsing the time strings using the corresponding formatter
DateTimeFormatter timeFormatter = DateTimeFormatter.ofPattern("H' hours : 'm' mins : 's' sec'");
for (String strTime : list) {
LocalTime time = LocalTime.parse(strTime, timeFormatter);
sumHours += time.getHour();
sumMinutes += time.getMinute();
sumSeconds += time.getSecond();
}
// Adjust hour, minutes and seconds if minute and/or second exceed 60
sumMinutes += sumSeconds / 60;
sumSeconds %= 60;
sumHours += sumMinutes / 60;
sumMinutes %= 60;
String strSum = String.format("%d hours : %d mins : %d sec", sumHours, sumMinutes, sumSeconds);
System.out.println(strSum);
}
}
25 hours : 12 mins : 33 sec
一种方法是使用。将每个时间值转换为持续时间,然后求和。提供的时间值以两个字符串的数组显示
String[] hours = {
"0 hours : 0 mins : 0 sec, 1 hours : 16 mins : 22 sec, 0 hours : 1 mins : 53"
+ "sec, 23 hours : 54 mins : 18 sec",
"0 hours : 8 mins : 22 sec, 0 hours : 8"
+ " mins : 22 sec, 0 hours : 8 mins : 22 sec" };
转换是使用流完成的。评论中解释了这一过程
List<Duration> times = Arrays.stream(hours)
// remove all the spaces in each string
.map(str -> str.replaceAll("\\s+", ""))
// split each string on the commas. At this
// point, each string is processed independently
// of the other in a separate stream
.map(str -> Arrays.stream(str.split(","))
// parse the time
.map(tm -> LocalTime.parse(tm,
DateTimeFormatter.ofPattern(
"H'hours':m'mins':s'sec'")))
// convert each time to a duration
.map(lt -> Duration
.between(LocalTime.of(0, 0, 0), lt))
// sum the durations
.reduce(Duration.ZERO, (a, b) -> a.plus(b)))
// and collect in a list
.collect(Collectors.toList());
印刷品
PT25H12M33S
PT25M6S
25 hours 12 minutes 33 seconds
25 minutes 6 seconds
要以更熟悉的格式打印它们,可以使用以下lambda
Function<Duration, String> durationToString = d -> {
StringBuilder timeString = new StringBuilder();
long h = d.toDaysPart()*24+d.toHoursPart();
timeString.append(h == 1 ? (h + " hour ") : h > 1 ? (h + " hours ") : "");
long m = d.toMinutesPart();
timeString.append(m == 1 ? (m + " minute ") : m > 1 ? (m + " minutes ") : "");
long s = d.toSecondsPart();
timeString.append(s == 1 ? (s + " second ") : s > 1 ? (s + " seconds ") : "");
return timeString.toString();
};
欢迎来到堆栈溢出。请学习如何使用堆栈溢出,并阅读如何提高问题的质量。然后,你的问题包括你的源代码作为一个,它可以被其他人编译和测试。另外,请检查以了解您可以提出哪些问题。请参阅:。请看:你的代码毫无意义。我看不出它有什么作用。您的内部循环迭代
shorter.size()
次,但您将shorter
初始化为空列表,然后再也不向其中添加任何内容,因此整个内部循环将永远不会运行。您的外部循环是关于在名为hours
的内容中进行迭代,您没有提供该内容的定义。但这并不重要,因为您只会将它的值分配给temp
,然后您就不会使用temp
<代码>毫秒似乎总是零。所以你的代码真的应该什么都不做。时间处理可能会变得出人意料的棘手,即使它一开始很简单,所以我建议尽量使用标准(经过测试和调试的)库。不幸的
25 hours 12 minutes 33 seconds
25 minutes 6 seconds