SimpleDateFormat解析返回错误的值 [英] SimpleDateFormat parse returns wrong value

问题描述

我有此代码:

public static String formatMinSecOrHourMinSec(final String length) {
        try {
            final SimpleDateFormat hhmmss = new SimpleDateFormat("HH:mm:ss", Locale.GERMAN);
            final Date date = hhmmss.parse(length);
            final GregorianCalendar gc0 = new GregorianCalendar(Locale.GERMAN);
            gc0.setTime(date);
            if(gc0.getTimeInMillis() >= 3600 * 1000){
                return hhmmss.format(gc0.getTime());
            }else{
                final SimpleDateFormat mmss = new SimpleDateFormat("mm:ss");
                return mmss.format(gc0.getTime());                    
            }
        } catch (final ParseException e) {
            LOGGER.debug("Konnte die Länge nicht parsen: " + length + "\n" + e);
            return length;
        }
    }

如果length设置为 01:29:00 ，我估计它将返回 01:29:00 ，但它返回 29:00 .这是因为gc0.getTimeInMillis()返回的时间比预期少一小时(3600 * 1000).我在做什么错了?

I estimate that it returns 01:29:00 if length is set to 01:29:00 but it returns 29:00. This is because gc0.getTimeInMillis() returns one hour less (3600 * 1000) than expected. What am I doing wrong ?

推荐答案

这是因为java.util.Date使用的是您的默认时区. (从date以毫秒为单位的打印时间，您将看到). 要解决此问题，请尝试:

this is because java.util.Date is using your default time zone. (print time in ms from date and you will see). To fix it try:

final SimpleDateFormat hhmmss = new SimpleDateFormat("HH:mm:ss");
hhmmss.setTimeZone(TimeZone.getTimeZone("UTC"));

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