分析前检查日期格式 [英] Check date format before parsing

问题描述

我正在使用字段持续时间解析几个文档。但是在不同的文件中，这是不同的格式，例如：

 持续时间：00:43
持续时间：113.046
持续时间：21.55 s
  

希望将它们全部解析为格式Duration：113.046，我怎么能在任何解析之前检查它是什么格式？

这段代码之前的一些条件，因为这是不正确的所有这些：

 持续时间长; 
 DateFormat sdf = new SimpleDateFormat（hh：mm：ss）; 
尝试{
日期durationD = sdf.parse（totalDuration）; 
日期zeroSec = sdf.parse（00:00:00）; 
 duration = durationD.getTime（） -  zeroSec.getTime（）; 
} catch（Exception e）{
 duration = Long.parseLong（totalDuration）; 
 
 
 
 $ b $ p 
 $ p 
 
 
 
 .htmlrel =nofollow noreferrer> regex 然后格式。这里有一个启发的例子： 
 
 
  Map< Pattern，DateFormat> dateFormatPatterns = new HashMap< Pattern，DateFormat>（）; 
 dateFormatPatterns.put（Pattern.compile（\\\\\ {1,2}：\\\\\\\\\\\\\\\\\\\\\\\\\\\\\' 
 dateFormatPatterns.put（Pattern.compile（\\\\\\\\\\\\\\\\\\\\\\\\' 
 dateFormatPatterns.put（Pattern.compile（\\\\\\\\\\\\\\\\\\' ）; 
 
 String [] strings = {00:43，113.046，21.55 s}; 
 DateFormat finalFormat = new SimpleDateFormat（HH：mm：ss）; （字符串string：字符串）{
 for（Pattern pattern：dateFormatPatterns.keySet（））{
 if（pattern.matcher（string）.matches（））{
 $ b $ 
 Date date = dateFormatPatterns.get（pattern）.parse（string）; 
 String formattedTime = finalFormat.format（date）; 
 System.out.println（formattedTime）; 
 break; 
 
 
 
 
 
 $ p 
 > 
 
 
 
 00:43:00 
 00:01:53 
 00:00:21 
 
 
I am parsing several documments with the field Duration. But in the differents files, it is in differnt formats, ex:
"Duration": "00:43"  
"Duration": "113.046"  
"Duration": "21.55 s"
I want to parse all of them to the format "Duration": "113.046", how could I check before any parsing in wich format it is??

Some conditions before this piece of code, because this is not right for all of them:
Long duration;
                    DateFormat sdf = new SimpleDateFormat("hh:mm:ss");
                    try {
                        Date durationD = sdf.parse(totalDuration);
                        Date zeroSec = sdf.parse("00:00:00");
                        duration = durationD.getTime() - zeroSec.getTime();
                    } catch (Exception e) {                     
                            duration = Long.parseLong(totalDuration);                       
                    }
Thanks in advance
解决方案
You could match the pattern with help of regex and then format accordingly. Here's a kickoff example:
Map<Pattern, DateFormat> dateFormatPatterns = new HashMap<Pattern, DateFormat>();
dateFormatPatterns.put(Pattern.compile("\\d{1,2}:\\d{2}"), new SimpleDateFormat("H:m"));
dateFormatPatterns.put(Pattern.compile("\\d{1,3}\\.\\d{3}"), new SimpleDateFormat("s.S"));
dateFormatPatterns.put(Pattern.compile("\\d{1,2}\\.\\d{2} s"), new SimpleDateFormat("s.S 's'"));

String[] strings = { "00:43", "113.046", "21.55 s" };
DateFormat finalFormat = new SimpleDateFormat("HH:mm:ss");

for (String string : strings) {
    for (Pattern pattern : dateFormatPatterns.keySet()) {
        if (pattern.matcher(string).matches()) {
            Date date = dateFormatPatterns.get(pattern).parse(string);
            String formattedTime = finalFormat.format(date);
            System.out.println(formattedTime);
            break;
        }
    }
}
This yields here
00:43:00
00:01:53
00:00:21


                        
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