java语言编程实现两个时间相差多少天、多少小时、多少分、多少秒
不多说,直接上干货!
DateDistance.java
package zhouls.bigdata.DataFeatureSelection.test; import java.text.DateFormat; import java.text.ParseException; import java.text.SimpleDateFormat; import java.util.Date; /** * 两个时间差计算 */ public class DateDistance { private static String distanceTime; public static void main(String[] args) throws Exception { String startday = "2017-09-20"; String endday ="2017-09-28"; long DistanceDays = getDistanceDays(startday,endday);//两个时间之间相差距离多少天 System.out.println(DistanceDays); String starttimes = "2017-01-17 00:10:20"; String endtimes ="2017-01-18 00:10:21"; long[] DistanceTimes = getDistanceTimes(starttimes,endtimes);//两个时间相差距离多少天多少小时多少分多少秒 ,以long[]形式返回 for (int i = 0; i < DistanceTimes.length; i++) { System.out.println(DistanceTimes[i]); } String DistanceTime = getDistanceTime(starttimes,endtimes);//两个时间相差距离多少天多少小时多少分多少秒 ,以String形式返回 System.out.println(DistanceTime); } /** * 两个时间之间相差距离多少天 * @param one 时间参数 1: * @param two 时间参数 2: * @return 相差天数 */ public static long getDistanceDays(String starttime, String endtime) throws Exception{ DateFormat df = new SimpleDateFormat("yyyy-MM-dd"); Date one; Date two; long days=0; try { one = df.parse(starttime); two = df.parse(endtime); long time1 = one.getTime(); long time2 = two.getTime(); long diff ; if(time1<time2) { diff = time2 - time1; } else { diff = time1 - time2; } days = diff / (1000 * 60 * 60 * 24); } catch (ParseException e) { e.printStackTrace(); } return days;//返回相差多少天 } /** * 两个时间相差距离多少天多少小时多少分多少秒 * @param str1 时间参数 1 格式:1990-01-01 12:00:00 * @param str2 时间参数 2 格式:2009-01-01 12:00:00 * @return long[] 返回值为:{天, 时, 分, 秒} */ public static long[] getDistanceTimes(String starttime, String endtime) { DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); Date one; Date two; long day = 0; long hour = 0; long min = 0; long sec = 0; try { one = df.parse(starttime); two = df.parse(endtime); long time1 = one.getTime(); long time2 = two.getTime(); long diff ; if(time1<time2) { diff = time2 - time1; } else { diff = time1 - time2; } day = diff / (24 * 60 * 60 * 1000); hour = (diff / (60 * 60 * 1000) - day * 24); min = ((diff / (60 * 1000)) - day * 24 * 60 - hour * 60); sec = (diff/1000-day*24*60*60-hour*60*60-min*60); } catch (ParseException e) { e.printStackTrace(); } long[] times = {day, hour, min, sec}; return times; } /** * 两个时间相差距离多少天多少小时多少分多少秒 * @param str1 时间参数 1 格式:1990-01-01 12:00:00 * @param str2 时间参数 2 格式:2009-01-01 12:00:00 * @return String 返回值为:xx天xx小时xx分xx秒 */ public static String getDistanceTime(String starttime, String endtime) { DateFormat df = new SimpleDateFormat("yyyy-MM-dd HH:mm:ss"); Date one; Date two; long day = 0; long hour = 0; long min = 0; long sec = 0; try { one = df.parse(starttime); two = df.parse(endtime); long time1 = one.getTime(); long time2 = two.getTime(); long diff ; if(time1<time2) { diff = time2 - time1; } else { diff = time1 - time2; } day = diff / (24 * 60 * 60 * 1000); hour = (diff / (60 * 60 * 1000) - day * 24); min = ((diff / (60 * 1000)) - day * 24 * 60 - hour * 60); sec = (diff/1000-day*24*60*60-hour*60*60-min*60); } catch (ParseException e) { e.printStackTrace(); } return day + "天" + hour + "小时" + min + "分" + sec + "秒"; } }
当然,我们可以进一步,将其放到一个时间的工具类去。
java编程如何实现多条2017-08-08 22:10:00.0这样的时间数据,相差多少天?(隔24小时为相差1天,否则为0天)
java编程如何实现多条2017-01-16 22:28:11.0这样的时间数据,转换成Date类型Mon Jan 16 22:28:11 CST 2017这样的时间数据
java编程如何实现2017-01-16 22:28:26.0这样的时间数据,转换成2017:01:16:22:28:26这样的时间数据
同时,这里大家也可以更改格式
DateDistance.java
package zhouls.bigdata.DataFeatureSelection.test; import java.text.DateFormat; import java.text.ParseException; import java.text.SimpleDateFormat; import java.util.Date; /** * 两个时间差计算 */ public class DateDistance { private static String distanceTime; public static void main(String[] args) throws Exception { String startday = "2017:09:20"; String endday ="2017:09:28"; long DistanceDays = getDistanceDays(startday,endday);//两个时间之间相差距离多少天 System.out.println(DistanceDays); String starttimes = "2017:01:17 00:10:20"; String endtimes ="2017:01:18 00:10:21"; long[] DistanceTimes = getDistanceTimes(starttimes,endtimes);//两个时间相差距离多少天多少小时多少分多少秒 ,以long[]形式返回 for (int i = 0; i < DistanceTimes.length; i++) { System.out.println(DistanceTimes[i]); } String DistanceTime = getDistanceTime(starttimes,endtimes);//两个时间相差距离多少天多少小时多少分多少秒 ,以String形式返回 System.out.println(DistanceTime); } /** * 两个时间之间相差距离多少天 * @param one 时间参数 1: * @param two 时间参数 2: * @return 相差天数 */ public static long getDistanceDays(String starttime, String endtime) throws Exception{ DateFormat df = new SimpleDateFormat("yyyy:MM:dd"); Date one; Date two; long days=0; try { one = df.parse(starttime); two = df.parse(endtime); long time1 = one.getTime(); long time2 = two.getTime(); long diff ; if(time1<time2) { diff = time2 - time1; } else { diff = time1 - time2; } days = diff / (1000 * 60 * 60 * 24); } catch (ParseException e) { e.printStackTrace(); } return days;//返回相差多少天 } /** * 两个时间相差距离多少天多少小时多少分多少秒 * @param str1 时间参数 1 格式:1990-01-01 12:00:00 * @param str2 时间参数 2 格式:2009-01-01 12:00:00 * @return long[] 返回值为:{天, 时, 分, 秒} */ public static long[] getDistanceTimes(String starttime, String endtime) { DateFormat df = new SimpleDateFormat("yyyy:MM:dd HH:mm:ss"); Date one; Date two; long day = 0; long hour = 0; long min = 0; long sec = 0; try { one = df.parse(starttime); two = df.parse(endtime); long time1 = one.getTime(); long time2 = two.getTime(); long diff ; if(time1<time2) { diff = time2 - time1; } else { diff = time1 - time2; } day = diff / (24 * 60 * 60 * 1000); hour = (diff / (60 * 60 * 1000) - day * 24); min = ((diff / (60 * 1000)) - day * 24 * 60 - hour * 60); sec = (diff/1000-day*24*60*60-hour*60*60-min*60); } catch (ParseException e) { e.printStackTrace(); } long[] times = {day, hour, min, sec}; return times; } /** * 两个时间相差距离多少天多少小时多少分多少秒 * @param str1 时间参数 1 格式:1990-01-01 12:00:00 * @param str2 时间参数 2 格式:2009-01-01 12:00:00 * @return String 返回值为:xx天xx小时xx分xx秒 */ public static String getDistanceTime(String starttime, String endtime) { DateFormat df = new SimpleDateFormat("yyyy:MM:dd HH:mm:ss"); Date one; Date two; long day = 0; long hour = 0; long min = 0; long sec = 0; try { one = df.parse(starttime); two = df.parse(endtime); long time1 = one.getTime(); long time2 = two.getTime(); long diff ; if(time1<time2) { diff = time2 - time1; } else { diff = time1 - time2; } day = diff / (24 * 60 * 60 * 1000); hour = (diff / (60 * 60 * 1000) - day * 24); min = ((diff / (60 * 1000)) - day * 24 * 60 - hour * 60); sec = (diff/1000-day*24*60*60-hour*60*60-min*60); } catch (ParseException e) { e.printStackTrace(); } return day + "天" + hour + "小时" + min + "分" + sec + "秒"; } }
作者:大数据和人工智能躺过的坑
出处:http://www.cnblogs.com/zlslch/
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