Java:如何解析日期格式以显示特定的输出格式?

4
在我的应用程序中,我以特定格式(由PHP生成)从数据库检索日期。
我想在我的Android应用程序中显示一种特定的输出格式,如下所示:
如果日期是今天,则我想以以下格式显示: "今天-17:50"
如果日期是昨天,则我想以以下格式显示: "昨天-17:50"
对于其他日期,我想以以下格式显示: "6月5日-17:50"
我该怎么做?
[更新]
String dateDebut = annonce.getDate_debut();

            SimpleDateFormat inDF = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); // inputFormat
            SimpleDateFormat TodayDF = new SimpleDateFormat("HH'h'mm"); //OutputFormat For today and yesterday
            SimpleDateFormat FullDF = new SimpleDateFormat("dd MMM - HH'h'mm"); //Outputformat long

            Date inDate = null;
            try {
                inDate = inDF.parse(dateDebut);
            } catch (ParseException e) {
                // TODO Auto-generated catch block
                e.printStackTrace();
            }
            //calendar for inputday
            Calendar inCal = new GregorianCalendar();
            inCal.setTime(inDate);
            //startOfToday
            Calendar cStartOfDate = new GregorianCalendar();
            cStartOfDate.set(Calendar.HOUR, 0);
            cStartOfDate.set(Calendar.MINUTE, 0);
            cStartOfDate.set(Calendar.SECOND, 0);
            cStartOfDate.set(Calendar.MILLISECOND, 0);
            //endOfToday    
            Calendar cEndOfDate = new GregorianCalendar();
            cEndOfDate.set(Calendar.HOUR, 23);
            cEndOfDate.set(Calendar.MINUTE, 59);
            cEndOfDate.set(Calendar.SECOND, 59);

             //startOfYesterday
            Calendar cStartOfYesterday = new GregorianCalendar();
            cStartOfYesterday.set(Calendar.HOUR, 0);
            cStartOfYesterday.set(Calendar.MINUTE, 0);
            cStartOfYesterday.set(Calendar.SECOND, 0);
            cStartOfYesterday.set(Calendar.MILLISECOND, 0);

             //endOfYesterday
            Calendar cEndOfYesterday = new GregorianCalendar();
            cEndOfYesterday.set(Calendar.HOUR, 23);
            cEndOfYesterday.set(Calendar.MINUTE, 59);
            cEndOfYesterday.set(Calendar.SECOND, 59);

            if (cStartOfDate.before(inCal) && cEndOfDate.after(inCal)){
              System.out.println("Aujourd'hui - "+TodayDF.format(inDate));
              viewHolder.dateDebut.setText("Aujourd'hui - "+TodayDF.format(inDate));
            } else if (cStartOfYesterday.before(inCal) && cEndOfYesterday.after(inCal)){
              System.out.println("Hier - "+TodayDF.format(inDate));
              viewHolder.dateDebut.setText("Hier - "+TodayDF.format(inDate));
            }  else {
              System.out.println(FullDF.format(inDate));
              viewHolder.dateDebut.setText(FullDF.format(inDate));
            }

2
你尝试了什么?你应该使用DateFormat(使用SimpleDateFormat实现)来处理5月6日 - 17:50这种格式:http://docs.oracle.com/javase/7/docs/api/java/text/SimpleDateFormat.html。对于“今天/昨天”,你可以使用`Calendar`(使用`GregorianCalendar`实现)。 - lpratlong
可能是如何在Java中计算“时间差”?的重复问题。 - Basil Bourque
4个回答

2

试一试这段代码:

  DateFormat inDF = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); // inputFormat
    DateFormat TodayDF = new SimpleDateFormat("HH'h'mm"); //OutputFormat For today and yesterday
    DateFormat FullDF = new SimpleDateFormat("dd MMM - HH'h'mm"); //Outputformat long

    Date inDate = inDF.parse("2014-06-05 17:50:50");
    //calendar for inputday
    Calendar inCal = new GregorianCalendar();
    inCal.setTime(inDate);
    //startOfToday
    Calendar cStartOfDate = new GregorianCalendar();
    cStartOfDate.set(Calendar.HOUR_OF_DAY, 0);
    cStartOfDate.set(Calendar.MINUTE, 0);
    cStartOfDate.set(Calendar.SECOND, 0);
    cStartOfDate.set(Calendar.MILLISECOND, 0);
    //endOfToday    
    Calendar cEndOfDate = new GregorianCalendar();
    cEndOfDate.set(Calendar.HOUR_OF_DAY, 23);
    cEndOfDate.set(Calendar.MINUTE, 59);
    cEndOfDate.set(Calendar.SECOND, 59);

     //startOfYesterday
    Calendar cStartOfYesterday = new GregorianCalendar();
    cStartOfYesterday.set(Calendar.HOUR_OF_DAY, 0);
    cStartOfYesterday.set(Calendar.MINUTE, 0);
    cStartOfYesterday.set(Calendar.SECOND, 0);
    cStartOfYesterday.set(Calendar.MILLISECOND, 0);

     //endOfYesterday
    Calendar cEndOfYesterday = new GregorianCalendar();
    cEndOfYesterday.set(Calendar.HOUR_OF_DAY, 23);
    cEndOfYesterday.set(Calendar.MINUTE, 59);
    cEndOfYesterday.set(Calendar.SECOND, 59);

    if (cStartOfDate.before(inCal) && cEndOfDate.after(inCal)){
      System.out.println("Today "+TodayDF.format(inDate));
    } else if (cStartOfYesterday.before(inCal) && cEndOfYesterday.after(inCal)){
      System.out.println("Yesterday"+TodayDF.format(inDate));
    }  else {

      System.out.println(FullDF.format(inDate));
    }

谢谢你的代码。但是似乎有问题,因为我只进入了else条件,从未显示今天和昨天。 - wawanopoulos
@wawanopoulos,你找到问题了吗?还是我来搜索一下? - Jens
@wawanopoulos 我已经再次测试过,对于以下示例,我得到的输出是 今天... - Jens
请看我在帖子中的更新。由于出现错误,我刚刚将代码中的DateFormat替换为SimpleDateFormat。并设置了我的输入日期。 - wawanopoulos
是的:一个数据块的示例:2014年05月30日17点50分50秒 - wawanopoulos
显示剩余2条评论

0
  • 首先将从数据库获取的日期转换为日历实例
  • 今天和昨天可以通过日历实例进行识别
  • 对于其他格式,请使用以下代码:

代码:

Calendar cal = Calendar.getInstance();
new SimpleDateFormat("dd MMMM - HH").format(cal.getTime())+ 
"h" + new SimpleDateFormat("mm").format(cal.getTime())

0

这里有一个完整的解决方案。我没有尝试过,但应该可以工作。

注意:要小心输入限制:例如,如果日期是01/01/2015,我不确定它是否能正常工作。 我让你来测试一下

private boolean checkSameDate(Calendar cal1, Calendar cal2) {
        if ((cal1.get(Calendar.YEAR) == cal2.get(Calendar.YEAR)) 
                && (cal1.get(Calendar.DAY_OF_YEAR) == cal2.get(Calendar.DAY_OF_YEAR))) {
            return true;
        }
        return false;
}

private void checkDate(Date date) {
        Calendar cal = new GregorianCalendar();
        cal.setTime(new Date());
        Calendar cal2 = new GregorianCalendar();
        cal2.setTime(date);
        cal.add(Calendar.DAY_OF_YEAR, 1);
        if (checkSameDate(cal, cal2)) {
            // Your input date is tomorrow.
        } else {
            cal.add(Calendar.DAY_OF_YEAR, -2);
            if (checkSameDate(cal, cal2)) {
                // Your input date is yesterday.
            } else {
                DateFormat sdf = new SimpleDateFormat("dd MMMM - HH:mm");
                System.out.println(sdf.format(date));
            }
        }
}

编辑

抱歉,我认为当今天是01/01/YYYY时,它不适用于31/12/YYYY-1的日期。也许您可以使用这种解决方案修复此代码:检查一个日期是否恰好在另一个日期之后24小时或更长时间

对于SimpleDateFormat,如果您需要此格式,请在此处查看https://ideone.com/dsxKN9

编辑2

我刚才看到您想要今天而不是明天:)。我的错!我会尝试修复这个问题,但是如果您理解了逻辑,就能够做到。


0
你可以使用这个函数来完成它:
public static String convertDate(String stringDate, String oldFormat) throws ParseException {        
        SimpleDateFormat sdf = new SimpleDateFormat(oldFormat);
        Date date = sdf.parse(stringDate);
        double daysAgo = (System.currentTimeMillis() - date.getTime()) / (24 * 60 * 60 * 1000d);
        System.out.println(daysAgo);
        String newFormat;
        if (daysAgo<=0){
            newFormat="'Today -' HH'h'mm";
        }
        else if (daysAgo>=0 && daysAgo<=1){
            newFormat="'Yesterday -' HH'h'mm";
        }
        else {
            newFormat="d MMMM '-' HH'h'mm";
        }
        sdf.applyPattern(newFormat);
        return sdf.format(date);
    }

使用方法如下:

String newDate = convertDate("2014-06-03 17:50:50", "yyyy-MM-dd HH:mm:ss");

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