正则表达式模式，用于从包含由定界符分隔的日期的字符串中删除.0

Question

I have a string which looks something like this:

"2013-10-19 22:21:21#2013-10-23 16:17:19#2013-10-25 13:15:14.0#2013-10-19 08:11:34.0#2013-10-23 16:17:19#"

I need to remove the .0 from the end of the date. ie 2013-10-25 13:15:14.0 should be 2013-10-25 13:15:14

What can be the possible regex pattern for this..

Answer 1

If that is your string and there's no other data inside, you can use a simple replace; no need for any regex:

String str = "2013-10-19 22:21:21#2013-10-23 16:17:19#2013-10-25 13:15:14.0#2013-10-19 08:11:34.0#2013-10-23 16:17:19#";
String res = str.replace(".0", "");
System.out.println(res);

Output:

2013-10-19 22:21:21#2013-10-23 16:17:19#2013-10-25 13:15:14#2013-10-19 08:11:34#2013-10-23 16:17:19# - See more at: http://ideone.com/WrDbqS#sthash.iQsWV8gp.dpuf

ideone demo

Answer 2

You can use:

str = str.replaceAll("\\.0(?=#)", "");

This will replace .0 with an empty string if it is followed by a #

Answer 3

当出现在#或字符串末尾时，将删除（用空白代替） .0 ：

str = str.replaceAll("\\.0(?=#|$)", "");

Answer 4

尝试这个

str = str.replaceAll("(\\d{4}-\\d{2}-\\d{2} \\d{2}:\\d{2}:\\d{2})\\.\\d+", "$1");

Answer 5

Without regex you can do as follows.

 String str="2013-10-19 22:21:21#2013-10-23 16:17:19#2013-10-25 13:15:14.0#2013-10-19 08:11:34.0#2013-10-23 16:17:19#";
   String[] dates=str.split("#");
   for(String i:dates){
       if(i.lastIndexOf('0')==(i.length()-1)){
           System.out.println(i.replaceAll(".0",""));
       }else {
           System.out.println(i);
       }
   }