java regex search and replace all?

Question

I have an example input that can be of any length and want to add a space every 4 characters.

for example input = xxxxxxxxxxxx

result I expect: xxxx xxxx xxxx

I have looked at the replaceAll() method but wondering how I can create a match that returns me the 4th, 8th, 12th etc character index so I can do something like this:

input.replaceAll("([\\S\\s)]{4})\\G", " " + input.charAt(index - 1))

where index somehow gets modified to get the appropriate index in which the regular expression of mine has found the 4th character.

How can it be done easier? I must note down that the input string is dynamic and can change. So while its being changed, i want to auto format it — Jono
– Jono, Commented Apr 30, 2015 at 14:05
I was thinking of Rahul Tripathi's answer, where you just parse the string and add a space every 4th character — Alexandru Severin
– Alexandru Severin, Commented Apr 30, 2015 at 14:11

Kent · Accepted Answer · 2015-04-30 14:05:22Z

4

"xxxxxxxxxxxx".replaceAll(".{4}(?!$)","$0 ");

This won't add trailing space to the last segment.

answered Apr 30, 2015 at 14:05

Kent

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5 Comments

Bohemian Over a year ago

.... is the same length, and easier to read, than .{4}, but +1 anyway, especially the "not at the end" whistle

Kent Over a year ago

@Bohemian thx. you are right for the ...., but if n>5, I would say .{n} would be easier to read. :)

Jono Over a year ago

It works if the format is already xxxxxxxxx But what if the format is only partially done ie xxxx xxxxxxxxxx? I tried that approach and what happens is that it increases the space from the first group of characters leading to something like this instead: xxxx xxxxxxxx with double space

Kent Over a year ago

@jonney want to add a space every 4 characters you didn't say anything about the "partially done" case. if you want to add space for every 4 non-empty chars, you can try \\S{4}(?!=\\s) I didn't test though.

Jono Over a year ago

@Kent any suggestions? tried the above you mentioned and still added extra space

Rahul Tripathi · Accepted Answer · 2015-04-30 14:04:24Z

2

You can try this wihout using regex:

StringBuilder str = new StringBuilder("xxxxxxxxxxxx");
int i = str.length() - 4;

while (i > 0)
{
    str.insert(i, " ");
    i = i - 4;
}

System.out.println(str.toString());

With regex:

String myString = "xxxxxxxxxxxx";
String str = myString.replaceAll("(.{4})(?!$)", "$0 ");
return str;

answered Apr 30, 2015 at 14:04

Rahul Tripathi

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Comments

Community · Accepted Answer · 2017-05-23 11:51:17Z

2

You can try with replaceAll(".{1,4}+(?!$)", "$0 ")

.{1,4}+ will match any 1-4 characters (+ will make it possessive)
(?!$) which are not right before end of string
$0 " will replace it with content from group 0 (which is current match) plus space

Actually I overcomplicated this regex. You can find simpler version based on same idea in @Kent's answer.

edited May 23, 2017 at 11:51

CommunityBot

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answered Apr 30, 2015 at 14:04

Pshemo

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Comments

Tassos Bassoukos · Accepted Answer · 2015-04-30 14:04:03Z

0

Use grouping; $1 (and upwards) will get replaced with matcher.getGroup(1) .

answered Apr 30, 2015 at 14:04

Tassos Bassoukos

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Comments

54l3d · Accepted Answer · 2015-04-30 14:32:44Z

0

This will not add a space if there is one

String s = "xxxxxxxx xxxxx x xx x";
System.out.println(s.replaceAll("\\w{4}(?!\\s)", "$0 " ));

Outmput: xxxx xxxx xxxx x x xx x

demo

edited Apr 30, 2015 at 14:32

answered Apr 30, 2015 at 14:23

54l3d

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