使用正则表达式删除结尾的特殊字符

Question

I need to remove trailing decimal/dot from a statement. Following is the statement

ABC paid 10.25 RS confirmed. from bank XYZ 20125722. on 23-12-2012

I want to remove the trailing decimals/dot from the statement. After removed the decimal/dot, it should be like

ABC paid 10.25 RS confirmed from bank XYZ 20125722 on 23-12-2012

Please note, I don't need to remove the decimal in the 10.25 , but in other places. How can I do it with Java?

Answer 1

在Java中，只需对字符串使用replaceAll（String pattern，String replacement）方法，如下所示：

String result = "ABC paid 10.25 RS confirmed. from bank XYZ 20125722. on 23-12-2012".replaceAll("\\.\\s+"," ") ;

Answer 2

The regex you could use is /\\.\\S+/ in Perl or Perl-regex supporting languages. Otherwise /\\.[ ^I]/ where "^I" is TAB character. Replace that with empty.

Hope that helps.

Answer 3

You should use is \\\\.\\\\s+ , because after the dot may contain many spaces or one space .

The code following:

String result = "ABC paid 10.25 RS confirmed. from bank  XYZ 20125722. on 23-12-2012"
               .replaceAll("\\.\\s+", " ");

Answer 4

So, you want to match any . not in a number? You can use a method called negative lookahead :

"\\.(?![0-9])"

This match any . NOT before a digit.

Answer 5

Your regex is

\.(?=\s|$)

means match a dot when it is followed by a whitespace or the end of the string. (?=\\s|$) is a positive lookahead that ensures that there is a whitespace or the end of the string following, but does not match this, so no need to insert it again, just replace with the empty string.

See it here on Regexr

For your java

String res = "ABC paid 10.25 RS confirmed. from bank XYZ 20125722. on 23-12-2012".replaceAll("\\.(?=\\s|$)", "");