我只是把这个一起很快,但类似:

static final String[] DAYS =
{ "Monday", "Tuesday", "Wednesday", "Thursday", "Friday", "Saturday", "Sunday" };

public class ActivityEvent
{
  public int day;
  public String name;
  public String typeOfExercise;
  public String time;
}

public List loadActivities(String filename) throws IOException
{
  List activities = new ArrayList();
  FileInputStream fis = new FileInputStream(filename);
  InputStreamReader isr = new InputStreamReader(fis);
  BufferedReader br = new BufferedReader(isr);
  int lastDay = -1;
  String line;
  while ((line = br.readLine()) != null)
  {
      line = line.trim();
      int day;
      for (day = DAYS.length - 1; day >= 0; day--)
      {
          if (line.equals(DAYS[day]))
          {
              break;
          }
      }
      String name;
      if (day < 0)
      {
          day = lastDay;
          if (lastDay < 0)
          {
              throw new IOException(filename + " must start with day of week");
          }
          name = line;
      }
      else
      {
          name = br.readLine();
          if (name == null)
          {
              throw new IOException(filename + " expected name, reached end of file");
          }
      }
      String type = br.readLine();
      if (type == null)
      {
          throw new IOException(filename + " expected type of exercise, reached end of file");
      }
      String time = br.readLine();
      if (time != null)
      {
          throw new IOException(filename + " expected time of exercise, reached end of file");
      }
      ActivityEvent activity = new ActivityEvent();
      activity.day = day;
      activity.name = name;
      activity.typeOfExercise = type;
      activity.time = time;
      activities.add(activity);
  }
  return activities;
}

public void printActivities(List activities)
{
    StringBuilder str = new StringBuilder("Day	Name	Type of Exercise	Time
");
    int numActivities = activities.size();
    int lastDay = -1;
    for (int index = 0; index < numActivities; index++)
    {
        ActivityEvent activity = (ActivityEvent)activities.get(index);
        if (activity.day != lastDay)
        {
            str.append(DAYS[activity.day]);
        }
        str.append( 	 );
        str.append(activity.name);
        str.append( 	 );
        str.append(activity.typeOfExercise);
        str.append( 	 );
        str.append(activity.time);
        str.append( 
 );
    }
    System.out.print(str.toString());
}

然后调用所有的例如:

List activities = loadActivities("somefile.txt");
// Do optional sorting, etc. here.
printActivities(activities);

我会看看Java s < a href = " http://java.sun.com/j2se/1.5.0/docs/api/java/io/PrintStream.html printf 28以%,% 20 Java . lang . object…% 29”rel = " nofollow noreferrer " > < / > sprintf()函数和它年代能力< a href = " http://www.java2s.com/Tutorial/Java/0120__Development/Rightjustifyingandleftjustifyingvalues.htm " rel = " nofollow noreferrer " >左/右证明< / >数据与指定的宽度。

<强>关于解析输入:< /强>

One issue you will have is that each "record" of data (each row, in the ouput) is not a fixed size. Some are 3-tuples of name,exercise,time, and others are 4-tuples of day,name,exercise,time

说,假设你已经给定的格式是真的都有,这个问题可以在工作。

After reading a line, you could check for a weekday, and if so assume that s the start of a 4-tuple, and read the next 3 lines. If it is not a weekday, then assume it is a 3-tuple, and only read the next 2 lines.

如果可能有“差距”的名称,类型,或时间列在输出中,在不同的组合,它将变得更加棘手。

你真的需要您的程序有特殊知识什么列中的值是有效的。如杰西卡不是一种有效的锻炼,慢跑并不是一个有效的名字。

<强>关于格式化输出< /强>

731952年< a href = " https://stackoverflow.com/questions/731917/how-to-display-data-from-txt-file-in-specific-format/731952 " >布莱恩回答< / >是相关的。

It depends on the language you use. Most languages have a printf-equivalent. The formatting codes of printf allow you to pad with space, etc.

如果您正在使用Perl(可能适合这个任务),您可以使用< a href = " http://www.perl.com/doc/manual/html/pod/perlform.html " rel = " nofollow noreferrer " > < / >格式