使用其他两个日期（提前三个小时和之后三个小时）进行比较。然后使用确认您要解析的日期在这两个边界之间
compareTo()。
您确实不应该在班级名称中输入数字，但这是一个与答案无关的样式问题。
public class SO31132861 {public static void main(String[] args) { SimpleDateFormat df = new SimpleDateFormat("yyyyMMddHHmmss"); df.setLenient(false); System.out.println(tryParse(df, "20160630231110")); System.out.println(tryParse(df, "20150228231100")); System.out.println(tryParse(df, "20160229231100")); System.out.println(tryParse(df, "21000229231100")); // 29th Feb on non-leap year 2100 System.out.println(tryParse(df, "20160631231110")); // 31st Jun invalid day System.out.println(tryParse(df, "20160229231160")); // Second > 59 System.out.println(tryParse(df, "20150229231100")); // 29th Feb on non-leap year 2015 System.out.println(tryParse(df, "20150228241100")); // Hour > 23}private static Boolean tryParse(DateFormat df, String s) { Boolean valid=false; try { Date threeHoursBefore = new Date(); threeHoursBefore.setTime(System.currentTimeMillis() - (3*60*60*1000)); Date threeHoursAfter = new Date(); threeHoursAfter.setTime(System.currentTimeMillis() + (3*60*60*1000)); Date dateToParse= df.parse(s); valid=dateToParse.compareTo(threeHoursBefore) > 0 && dateToParse.compareTo(threeHoursAfter) < 0; } catch (ParseException e) { valid=false; } return valid;}}
