java - 如何在 Java 中转换 Youtube API V3 持续时间

Question

Youtube V3 API 使用 ISO8601 时间格式来描述视频的时长。像“PT1M13S”这样的东西。现在我想将字符串转换为秒数（例如本例中的 73）。

是否有任何 Java 库可以帮助我轻松完成 Java 6 下的任务？还是我必须自己完成正则表达式任务？

编辑

最后我接受@Joachim Sauer 的回答

示例代码Joda如下。

PeriodFormatter formatter = ISOPeriodFormat.standard();
Period p = formatter.parsePeriod("PT1H1M13S");
Seconds s = p.toStandardSeconds();

System.out.println(s.getSeconds());

score 15 · Accepted Answer

Joda Time是任何类型的时间相关函数的首选库。

对于这种特定情况， ISOPeriodFormat.standard()返回一个PeriodFormatter可以解析和格式化该格式的值。

结果对象是(Period JavaDoc )。获得实际的秒数将是period.toStandardSeconds().getSeconds()，但我建议您将持续时间作为一个Period对象处理（为了便于处理和类型安全）。

编辑：来自未来我的便条：这个答案已经有好几年了。Java 8 带来了java.time.Duration它也可以解析这种格式并且不需要外部库。

score 13 · Accepted Answer

13

Java 8 中的解决方案：

Duration.parse(duration).getSeconds()

于 2014-11-12T09:55:56.130 回答

score 13 · Accepted Answer

我可能会迟到，其实很简单。尽管可能有更好的方法来做到这一点。持续时间以毫秒为单位。

public long getDuration() {
    String time = "PT15H12M46S".substring(2);
    long duration = 0L;
    Object[][] indexs = new Object[][]{{"H", 3600}, {"M", 60}, {"S", 1}};
    for(int i = 0; i < indexs.length; i++) {
        int index = time.indexOf((String) indexs[i][0]);
        if(index != -1) {
            String value = time.substring(0, index);
            duration += Integer.parseInt(value) * (int) indexs[i][1] * 1000;
            time = time.substring(value.length() + 1);
        }
    }
    return duration;
}

score 3 · Accepted Answer

这是我的解决方案

public class MyDateFromat{
    public static void main(String args[]){
        String ytdate = "PT1H1M15S";
        String result = ytdate.replace("PT","").replace("H",":").replace("M",":").replace("S","");
        String arr[]=result.split(":");
        String timeString = String.format("%d:%02d:%02d", Integer.parseInt(arr[0]), Integer.parseInt(arr[1]),Integer.parseInt(arr[2]));
        System.out.print(timeString);       
    }
}

如果您想在几秒钟内转换，它将返回 H:MM:SS 格式的字符串，您可以使用

int timeInSedonds = int timeInSecond = Integer.parseInt(arr[0])*3600 + Integer.parseInt(arr[1])*60 +Integer.parseInt(arr[2])

注意：可能会抛出异常，请根据 result.split(":"); 的大小处理。

score 1 · Accepted Answer

可能这会帮助一些不想要任何库但一个简单功能的人，

String duration="PT1H11M14S";

这是功能，

private String getTimeFromString(String duration) {
    // TODO Auto-generated method stub
    String time = "";
    boolean hourexists = false, minutesexists = false, secondsexists = false;
    if (duration.contains("H"))
        hourexists = true;
    if (duration.contains("M"))
        minutesexists = true;
    if (duration.contains("S"))
        secondsexists = true;
    if (hourexists) {
        String hour = "";
        hour = duration.substring(duration.indexOf("T") + 1,
                duration.indexOf("H"));
        if (hour.length() == 1)
            hour = "0" + hour;
        time += hour + ":";
    }
    if (minutesexists) {
        String minutes = "";
        if (hourexists)
            minutes = duration.substring(duration.indexOf("H") + 1,
                    duration.indexOf("M"));
        else
            minutes = duration.substring(duration.indexOf("T") + 1,
                    duration.indexOf("M"));
        if (minutes.length() == 1)
            minutes = "0" + minutes;
        time += minutes + ":";
    } else {
        time += "00:";
    }
    if (secondsexists) {
        String seconds = "";
        if (hourexists) {
            if (minutesexists)
                seconds = duration.substring(duration.indexOf("M") + 1,
                        duration.indexOf("S"));
            else
                seconds = duration.substring(duration.indexOf("H") + 1,
                        duration.indexOf("S"));
        } else if (minutesexists)
            seconds = duration.substring(duration.indexOf("M") + 1,
                    duration.indexOf("S"));
        else
            seconds = duration.substring(duration.indexOf("T") + 1,
                    duration.indexOf("S"));
        if (seconds.length() == 1)
            seconds = "0" + seconds;
        time += seconds;
    }
    return time;
}

score 0 · Accepted Answer

您可以使用标准SimpleDateFormat解析String到 aDate并从那里处理它：

DateFormat df = new SimpleDateFormat("'PT'mm'M'ss'S'");
String youtubeDuration = "PT1M13S";
Date d = df.parse(youtubeDuration);
Calendar c = new GregorianCalendar();
c.setTime(d);
c.setTimeZone(TimeZone.getDefault());
System.out.println(c.get(Calendar.MINUTE));
System.out.println(c.get(Calendar.SECOND));

score 0 · Accepted Answer

如果您可以确定输入的有效性并且不能使用正则表达式，我使用此代码（以毫秒为单位返回）：

Integer parseYTDuration(char[] dStr) {
    Integer d = 0;

    for (int i = 0; i < dStr.length; i++) {
        if (Character.isDigit(dStr[i])) {
            String digitStr = "";
            digitStr += dStr[i];
            i++;
            while (Character.isDigit(dStr[i])) {
                digitStr += dStr[i];
                i++;
            }

            Integer digit = Integer.valueOf(digitStr);

            if (dStr[i] == 'H')
                d += digit * 3600;
            else if (dStr[i] == 'M')
                d += digit * 60;
            else
                d += digit;
        }
    }

    return d * 1000;
}

score 0 · Accepted Answer

我已经实现了这个方法，到目前为止它已经奏效了。

private String timeHumanReadable (String youtubeTimeFormat) {
// Gets a PThhHmmMssS time and returns a hh:mm:ss time

    String
            temp = "",
            hour = "",
            minute = "",
            second = "",
            returnString;

    // Starts in position 2 to ignore P and T characters
    for (int i = 2; i < youtubeTimeFormat.length(); ++ i)
    {
        // Put current char in c
        char c = youtubeTimeFormat.charAt(i);

        // Put number in temp
        if (c >= '0' && c <= '9')
            temp = temp + c;
        else
        {
            // Test char after number
            switch (c)
            {
                case 'H' : // Deal with hours
                    // Puts a zero in the left if only one digit is found
                    if (temp.length() == 1) temp = "0" + temp;

                    // This is hours
                    hour = temp;

                    break;

                case 'M' : // Deal with minutes
                    // Puts a zero in the left if only one digit is found
                    if (temp.length() == 1) temp = "0" + temp;

                    // This is minutes
                    minute = temp;

                    break;

                case  'S': // Deal with seconds
                    // Puts a zero in the left if only one digit is found
                    if (temp.length() == 1) temp = "0" + temp;

                    // This is seconds
                    second = temp;

                    break;

            } // switch (c)

            // Restarts temp for the eventual next number
            temp = "";

        } // else

    } // for

    if (hour == "" && minute == "") // Only seconds
        returnString = second;
    else {
        if (hour == "") // Minutes and seconds
            returnString = minute + ":" + second;
        else // Hours, minutes and seconds
            returnString = hour + ":" + minute + ":" + second;
    }

    // Returns a string in hh:mm:ss format
    return returnString; 

}

score 0 · Accepted Answer

我自己做的

我们试试看

import java.text.DateFormat;
import java.text.ParseException;
import java.text.SimpleDateFormat;
import java.util.Calendar;
import java.util.Date;
import java.util.GregorianCalendar;
import java.util.

public class YouTubeDurationUtils {
    /**
     * 
     * @param duration
     * @return "01:02:30"
     */
    public static String convertYouTubeDuration(String duration) {
        String youtubeDuration = duration; //"PT1H2M30S"; // "PT1M13S";
        Calendar c = new GregorianCalendar();
        try {
            DateFormat df = new SimpleDateFormat("'PT'mm'M'ss'S'");
            Date d = df.parse(youtubeDuration);
            c.setTime(d);
        } catch (ParseException e) {
            try {
                DateFormat df = new SimpleDateFormat("'PT'hh'H'mm'M'ss'S'");
                Date d = df.parse(youtubeDuration);
                c.setTime(d);
            } catch (ParseException e1) {
                try {
                    DateFormat df = new SimpleDateFormat("'PT'ss'S'");
                    Date d = df.parse(youtubeDuration);
                    c.setTime(d);
                } catch (ParseException e2) {
                }
            }
        }
        c.setTimeZone(TimeZone.getDefault());

        String time = "";
        if ( c.get(Calendar.HOUR) > 0 ) {
            if ( String.valueOf(c.get(Calendar.HOUR)).length() == 1 ) {
                time += "0" + c.get(Calendar.HOUR);
            }
            else {
                time += c.get(Calendar.HOUR);
            }
            time += ":";
        }
        // test minute
        if ( String.valueOf(c.get(Calendar.MINUTE)).length() == 1 ) {
            time += "0" + c.get(Calendar.MINUTE);
        }
        else {
            time += c.get(Calendar.MINUTE);
        }
        time += ":";
        // test second
        if ( String.valueOf(c.get(Calendar.SECOND)).length() == 1 ) {
            time += "0" + c.get(Calendar.SECOND);
        }
        else {
            time += c.get(Calendar.SECOND);
        }
        return time ;
    }
}

score 0 · Accepted Answer

还有另一个很长的路要走。

// PT1H9M24S -->  1:09:24
// PT2H1S"   -->  2:00:01
// PT23M2S   -->  23:02
// PT31S     -->  0:31

public String convertDuration(String duration) {
    duration = duration.substring(2);  // del. PT-symbols
    String H, M, S;
    // Get Hours:
    int indOfH = duration.indexOf("H");  // position of H-symbol
    if (indOfH > -1) {  // there is H-symbol
        H = duration.substring(0,indOfH);      // take number for hours
        duration = duration.substring(indOfH); // del. hours
        duration = duration.replace("H","");   // del. H-symbol
    } else {
        H = "";
    }
    // Get Minutes:
    int indOfM = duration.indexOf("M");  // position of M-symbol
    if (indOfM > -1) {  // there is M-symbol
        M = duration.substring(0,indOfM);      // take number for minutes
        duration = duration.substring(indOfM); // del. minutes
        duration = duration.replace("M","");   // del. M-symbol
        // If there was H-symbol and less than 10 minutes
        // then add left "0" to the minutes
        if (H.length() > 0 && M.length() == 1) {
            M = "0" + M;
        }
    } else {
        // If there was H-symbol then set "00" for the minutes
        // otherwise set "0"
        if (H.length() > 0) {
            M = "00";
        } else {
            M = "0";
        }
    }
    // Get Seconds:
    int indOfS = duration.indexOf("S");  // position of S-symbol
    if (indOfS > -1) {  // there is S-symbol
        S = duration.substring(0,indOfS);      // take number for seconds
        duration = duration.substring(indOfS); // del. seconds
        duration = duration.replace("S","");   // del. S-symbol
        if (S.length() == 1) {
            S = "0" + S;
        }
    } else {
        S = "00";
    }
    if (H.length() > 0) {
        return H + ":" +  M + ":" + S;
    } else {
        return M + ":" + S;
    }
}

score 0 · Accepted Answer

我已经编写并使用此方法来获取实际持续时间。希望这可以帮助。

private String parseDuration(String duration) {
    duration = duration.contains("PT") ? duration.replace("PT", "") : duration;
    duration = duration.contains("S") ? duration.replace("S", "") : duration;
    duration = duration.contains("H") ? duration.replace("H", ":") : duration;
    duration = duration.contains("M") ? duration.replace("M", ":") : duration;
    String[] split = duration.split(":");
    for(int i = 0; i< split.length; i++){
        String item = split[i];
        split[i] = item.length() <= 1 ? "0"+item : item;
    }
    return TextUtils.join(":", split);
}

score 0 · Accepted Answer

时间已经被格式化，所以似乎只需替换就足够了。

 private String stringForTime(String ytFormattedTime) {
                 return ytFormattedTime
                        .replace("PT","")
                        .replace("H",":")
                        .replace("M",":")
                        .replace("S","");
 }

score 0 · Accepted Answer

public String pretty(String duration) {
  String time = duration.replace("PT", "");
  String hour = null;
  String minute = null;
  String second = null;

  if (time.indexOf("H") > 0) {
    String[] split = time.split("H");
    if (split.length > 0) {
      hour = split[0];
    }
    if (split.length > 1) {
      time = split[1];
    }
  }

  if (time.indexOf("M") > 0) {
    String[] split = time.split("M");
    if (split.length > 0) {
      minute = split[0];
    }
    if (split.length > 1) {
      time = split[1];
    }
  }

  if (time.indexOf("S") > 0) {
    String[] split = time.split("S");
    if (split.length > 0) {
      second = split[0];
    }
  }

  if (TextUtils.isEmpty(hour)) {
    if (TextUtils.isEmpty(minute)) { return "0:" + pad(second, 2, '0'); }
    else { return minute + ":" + pad(second, 2, '0'); }
  }
  else {
    if (TextUtils.isEmpty(minute)) { return hour + ":00:" + pad(second, 2, '0'); }
    else {return hour + ":" + pad(minute, 2, '0') + ":" + pad(second, 2, '0');}
  }
}

private String pad(String word, int length, char ch) {
  if (TextUtils.isEmpty(word)) { word = ""; }
  return length > word.length() ? pad(ch + word, length, ch) : word;
}

score -1 · Accepted Answer

Converting ISO 8601-compliant String to java.util.Date问题包含另一个解决方案：

更简单的解决方案可能是使用 JAXB 中的数据类型转换器，因为 JAXB 必须能够根据 XML Schema 规范解析 ISO8601 日期字符串。 javax.xml.bind.DatatypeConverter.parseDateTime("2010-01-01T12:00:00Z") 会给你一个对象，如果你需要一个对象Calendar，你可以简单地使用它。getTime()Date

score -1 · Accepted Answer

使用本网站：

// URL that generated this code:
// http://txt2re.com/index-java.php3?s=PT1M13S&6&3&18&20&-19&-21 

import java.util.regex.*;

class Main
{
  public static void main(String[] args)
  {
    String txt="PT1M13S";

    String re1="(P)";   // Any Single Character 1
    String re2="(T)";   // Any Single Character 2
    String re3="(\\d+)";    // Integer Number 1
    String re4="(M)";   // Any Single Character 3
    String re5="(\\d+)";    // Integer Number 2
    String re6="(S)";   // Any Single Character 4

    Pattern p = Pattern.compile(re1+re2+re3+re4+re5+re6,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
    Matcher m = p.matcher(txt);
    if (m.find())
    {
        String c1=m.group(1);
        String c2=m.group(2);
        String minutes=m.group(3); // Minutes are here
        String c3=m.group(4);
        String seconds=m.group(5); // Seconds are here
        String c4=m.group(6);
        System.out.print("("+c1.toString()+")"+"("+c2.toString()+")"+"("+minutes.toString()+")"+"("+c3.toString()+")"+"("+seconds.toString()+")"+"("+c4.toString()+")"+"\n");

        int totalSeconds = Integer.parseInt(minutes) * 60 + Integer.parseInt(seconds);
    }
  }
}

java - 如何在 Java 中转换 Youtube API V3 持续时间

15 回答 15

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