存储为字符串的两次之间的差异[重复]

Question

我有两个格式为“dd-MM-YYYY HH:mm:ss”的字符串。如何找到两个字符串之间的小时差？

Answer 1

您可以将String 转换为Date 对象，然后您可以比较它们。 类似下面的东西

 SimpleDateFormat sdformat = new SimpleDateFormat("dd-MM-YYYY HH:mm:ss");
  Date d1 = sdformat.parse("04-05-2020 02:33:12");
  Date d2 = sdformat.parse("05-05-2020 02:33:12");
  System.out.println("The date 1 is: " + sdformat.format(d1));
  System.out.println("The date 2 is: " + sdformat.format(d2));
  if(d1.compareTo(d2) > 0) {
     System.out.println("Date 1 occurs after Date 2");
  } else if(d1.compareTo(d2) < 0) {
     System.out.println("Date 1 occurs before Date 2");
  } else if(d1.compareTo(d2) == 0) {
     System.out.println("Both dates are equal");
  }

Answer 2

您可以使用SimpleDateFormat 和Date：

import java.text.ParseException; 
import java.text.SimpleDateFormat; 
import java.util.Date; 


public static long date_hour_diff(String start_date, String end_date) { 
  SimpleDateFormat simple_date_format = new SimpleDateFormat("dd-MM-yyyy HH:mm:ss"); 
  try {  
    Date parsed_start_date = simple_date_format.parse(start_date); 
    Date parsed_end_date = simple_date_format.parse(end_date); 
  
    long time_diff = d2.getTime() - d1.getTime(); 
    long diff_hours = (time_diff / (1000 * 60 * 60)) % 24; 
    return diff_hours;
  } catch (ParseException e) { 
      Log.e(TAG, "Date is not valid.");
  } 
}