[英]Check if a given time-stamp lies between two time-stamp in android
下面是我的代碼:
public static boolean isTimeBetweenTwoTime(String initialTime, String finalTime, String currentTime) throws ParseException {
String reg = "^([0-1][0-9]|2[0-3]):([0-5][0-9]):([0-5][0-9])$";
if (initialTime.matches(reg) && finalTime.matches(reg) && currentTime.matches(reg)) {
boolean valid = false;
//Start Time
java.util.Date inTime = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse(initialTime);
Calendar calendar1 = Calendar.getInstance();
calendar1.setTime(inTime);
//Current Time
java.util.Date checkTime = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse(currentTime);
Calendar calendar3 = Calendar.getInstance();
calendar3.setTime(checkTime);
//End Time
java.util.Date finTime = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse(finalTime);
Calendar calendar2 = Calendar.getInstance();
calendar2.setTime(finTime);
if (finalTime.compareTo(initialTime) < 0) {
calendar2.add(Calendar.DATE, 1);
calendar3.add(Calendar.DATE, 1);
}
java.util.Date actualTime = calendar3.getTime();
if ((actualTime.after(calendar1.getTime()) || actualTime.compareTo(calendar1.getTime()) == 0)&& actualTime.before(calendar2.getTime())) {
valid = true;
}
return valid;
} else {
throw new IllegalArgumentException("Not a valid time, expecting MM/dd/yyyy HH:mm:ss format");
}
}
但是它對我不起作用,請幫忙
試試這個簡單的邏輯:
long mills = System.currentTimeMillis();
SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss");
Date resultdate = new Date(mills);
String currentTime = sdf.format(resultdate);
System.out.println(sdf.format(resultdate));
try{
java.util.Date inTime1 = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse(initialTime);
java.util.Date inTime2 = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse(finalTime);
java.util.Date inTime3 = new SimpleDateFormat("MM/dd/yyyy HH:mm:ss").parse(currentTime);
if (inTime3.getTime() > inTime1.getTime() && inTime3.getTime() < inTime2.getTime()){
Log.e("TimeDifference","inTime3 is between inTime1 and inTime2");
return true;
}else{
Log.e("TimeDifference","in Else Condition");
return false;
}
}catch (Exception e){
e.printStackTrace();
return false;
}
請告訴您是否有任何問題。
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