I have strings with values "Address Line1", "Address Line2" ... etc. I want to add a space if there is any numeric value in the string like "Address Line 1", "Address Line 2".
I can do this using contains and replace like this
String sample = "Address Line1";
if (sample.contains("1")) {
sample = sample.replace("1"," 1");
}
But how can I do this using regex?
sample = sample.replaceAll("\\d+"," $0");
To use regex you will need replaceAll instead of replace method:
as regex you can use
\\d+ to match any group of one or more continues digits. We need all continues digits here because matching only one would create from foo123 something like foo 1 2 3(?<=[a-zA-Z])\\d if you want to add space only before digit which has alphabetic character before it. (?<=\\[a-zA-Z]) part is look-behind and it just checks if tested digit has character from range a-z or A-Z before it.and as replacement you can use " $0 which means space and match from group 0 which means part currently matched by regex.
So try with
sample = sample.replaceAll("\\d+", " $0")
or
sample = sample.replaceAll("(?<=[a-zA-Z])\\d", " $0")
which will change "hello 1 world2" into "hello 1 world 2" - notice that only 2 has additional space.
$0 is reference to group 0 which represents match from entire regex. I don't need any additional groups here.First Create a Pattern Object of what you want to search and compile it in your case Pattern object will be as follows:-
Pattern p=Pattern.compile("1");
Now Create Matcher object for your string
Matcher m=p.matcher(sample);
Now put a condition to check if Matcher has found any your Pattern String and if it has put a replaceAll method to replace it
if(m.find())
{
sample=m.replaceAll(" 1");
}
The Complete code is as follows:-
import java.io.*;
import java.util.regex.*;
class demo
{
public static void main(String args[])
{
String sample = "Address Line1";
Pattern p=Pattern.compile("1");
Matcher m=p.matcher(sample);
if(m.find())
{
sample=m.replaceAll(" 1");
}
System.out.println(sample);
}
}
