String type time comparison method with error string.compareto

Recently because of a bug Look at the old projects of the company , Found the following code ：

		/** *  Returns the number of seconds between , Get the settings within the specified time , If not within the specified time , Return to null * @param conf  Configuration information class  * @param time  At present, it is HH:mm:ss, And start and end Not unified ,start and end yes HH:mm * @return */
        private String getDistanceSecond(RepostUser conf,String time)
        {
    
        	if (conf.getStart1()!=null && !conf.getStart1().equals("") && time.compareTo(conf.getStart1()) >= 0 && time.compareTo(conf.getEnd1()) <= 0)
        	{
    
        		return  (conf.getDistanceSecond()+ conf.getDistanceTime()*60)+","+conf.getNum();
        	}
        	else if (conf.getStart2() != null && conf.getStart2() != null && !conf.getStart2().equals("") && time.compareTo(conf.getStart2()) >= 0 && time.compareTo(conf.getEnd2()) <= 0)
        	{
    
        		return  conf.getDistanceSecond2()+ conf.getDistanceTime2()*60+","+conf.getNum2();
        	}
        	else if (conf.getStart3()!=null && conf.getStart3() != null && !conf.getStart3().equals("") && time.compareTo(conf.getStart3()) >= 0 && time.compareTo(conf.getEnd3()) <= 0)
        	{
    
        		return  conf.getDistanceSecond3()+ conf.getDistanceTime3()*60+","+conf.getNum3();
        	}
        	else if (conf.getStart4() != null && conf.getStart4() != null && !conf.getStart4().equals("") && time.compareTo(conf.getStart4()) >= 0 && time.compareTo(conf.getEnd4()) <= 0)
        	{
    
        		return  conf.getDistanceSecond4()+ conf.getDistanceTime4()*60+","+conf.getNum4();
        	}
        	else if (conf.getStart5()!=null && conf.getStart5() != null &&  !conf.getStart5().equals("") && time.compareTo(conf.getStart5()) >= 0 && time.compareTo(conf.getEnd5()) <= 0)
        	{
    
        		return  conf.getDistanceSecond5()+conf.getDistanceTime5()*60+","+conf.getNum5();
        	}
        	return  null;// Indicates that it is out of range 
        }

there start and end, The format is HH:mm
time It is HH:mm:ss
here start and end , Why and time The format is different , I don't care , Just as start and end The number of seconds is 00 To deal with .
According to the above , I wrote a test program .

public class MainApp {
    
    public static void main(String[] args) {
    
        String time="16:40:00";
        String start="16:35";
        String end="16:40";
        if(time.compareTo(start) >= 0 && time.compareTo(end) <= 0){
    
            System.out.println(" In the interval ");
        }else{
    
            System.out.println(" Not in the interval ");
        }
    }
}

Results output ：

OK, Then say so , The original code does not contain boundaries ？ Then I'll try again time=16:35:00

public class MainApp {
    
    public static void main(String[] args) {
    
        String time="16:35:00";
        String start="16:35";
        String end="16:40";
        if(time.compareTo(start) >= 0 && time.compareTo(end) <= 0){
    
            System.out.println(" In the interval ");
        }else{
    
            System.out.println(" Not in the interval ");
        }
    }
}

Results output ：

In the interval

！！！, Then this is wrong , The boundary either contains , Or it doesn't include , What do you mean by giving me a include and a exclude ？

Actually, this one uses String Of compareTo Method to compare , This method is like this ：

    public int compareTo(String anotherString) {
    
        int len1 = value.length;
        int len2 = anotherString.value.length;
        int lim = Math.min(len1, len2);
        char v1[] = value;
        char v2[] = anotherString.value;

        int k = 0;
        while (k < lim) {
    
            char c1 = v1[k];
            char c2 = v2[k];
            if (c1 != c2) {
    
                return c1 - c2;
            }
            k++;
        }
        return len1 - len2;
    }

First compareTo Will compare each character of the string ASCII The size of a yard , If the range , Then return the difference between them . Otherwise, continue to find unequal characters .
If in Math.min(len1,len2) Characters within , It's all the same .
Then directly return the difference of their string length .

So I think this method can compare the size of time ？ My answer is not necessarily , If they are all the same length , In theory, it's really no problem , But you make me grow differently , I can do nothing about it, either , We have to use a more accurate time to compare .
So I use JDK8 Time provided api（ stay java.time Inside the bag ） refactoring .

public class MainApp {
    
    public static void main(String[] args) {
    
        String time="16:35:00";
        String start="16:35";
        String end="16:40";
        

        LocalTime time2=LocalTime.parse(time,DateTimeFormatter.ofPattern("HH:mm:ss"));
        LocalTime time3=LocalTime.parse(start,DateTimeFormatter.ofPattern("HH:mm"));
        LocalTime time4=LocalTime.parse(end,DateTimeFormatter.ofPattern("HH:mm"));
        System.out.println(time3);

        if (time2.compareTo(time3) >= 0 && time2.compareTo(time4) <= 0) {
    
            System.out.println(" In the interval ");
        }else{
    
            System.out.println(" Not in the interval ");
        }
    }
}

In this way, it will be more in line with normal thinking .
But if it's too Many objects are generated , I think we can take the evil route . Direct will start and end Add... To the back of :00 Just fine .

public class MainApp {
    
    public static void main(String[] args) {
    
        String time="16:40:00";
        String start="16:35";
        String end="16:40";

        if(start.length()==5){
    
            start+=":00";
        }
        if(end.length()==5){
    
            end+=":00";
        }
        if (time.compareTo(start) >= 0 && time.compareTo(end) <= 0) {
    
            System.out.println(" In the interval ");
        }else{
    
            System.out.println(" Not in the interval ");
        }

It's also possible to , Because the larger the number ,ASCII The bigger the size , In the case of equal length , It can be more correct .