Replace special chars with regex Java

Question

Looking for some help to find how to replace the last "." char to some other string via regex in java. I have code something like this which doesn't work. Any advise.

"//myserver.com//NAGA.CSV".replaceAll(".$","_2.")

Output: //myserver.com//NAGA_2.CSV

[`"//myserver.com//NAGA.CSV".replaceFirst("""\.(?!.*\.)""","_2.")`](https://ideone.com/JTalEH) — Wiktor Stribiżew, Aug 07 '17 at 10:32
Also [Replace last part of string](https://stackoverflow.com/questions/1660034/replace-last-part-of-string) — Flown, Aug 07 '17 at 10:32

score 3 · Accepted Answer · answered Aug 07 '17 at 10:31

You could use a lookahead, to find the end of the string, preceded by characters other than dot.

yourString.replaceAll("\\.(?=[^.]*$)", "replacement");

Note that the first dot needs to be escaped with a backslash, because dot has a special meaning in a regular expression (it matches any character). The second dot doesn't need to be escaped, because the special meaning doesn't apply in square brackets.

The (?= ) structure means "followed by this" - in other words, the dot that you match can be followed by any number of non-dot characters, and then the end of the string. Those extra characters are not considered part of the match.

score 0 · Answer 2 · answered Aug 07 '17 at 11:14

It's not a regex, but you could use the String.lastIndexOf() method to get the position of the last occurence of your char and use String.substring to create your new String:

String yourString = "test.csv";
String newValue = "_2.";
int lastOccurence = yourString.lastIndexOf(".");
String replacedString = yourString.substring(0, lastOccurence) + newValue + yourString.substring(lastOccurence + 1);

Replace special chars with regex Java

2 Answers2