Java regex and replacement

Question

Hi I am trying to understand Java regex replacement. I have lots of regex and replacement to apply on text in a file. I want to read regex and apply replacement on text. Like, I want to replace text to variable in following example.

import java.util.regex.*;
public class regex1{
public static void main(String args[]){
    String s1 = "cat catches dog text";
    Pattern p1 = Pattern.compile("\\s*cat\\s+catches\\s*dog\\s+(\\S+)");
    Matcher m1 = p1.matcher(s1);
    if (m1.find()){
        System.out.println(m1.group(1));
        s1 = m1.replaceFirst("variable $1");
        System.out.println(s1);
    }
    else{
        System.out.println("Else");
    }
}    
}

But I get output as

text
variable text

Can any one explain how does group and replacement works in java? How to get correct output?

If you just want to replace text with variable in your string, you don't need regex. If you want to do more than that, then show the real code. — Rohit Jain
– Rohit Jain, Commented Jul 8, 2013 at 18:21
You only have one group at the end of the expression which matches "text". So there is only one group :-) (or the whole match (0)) — Bart
– Bart, Commented Jul 8, 2013 at 18:21
@Bart yes i got that, my understanding is it will replace group 1 text to variable. But it is not happening like that. What is correct regex ? — Netro
– Netro, Commented Jul 8, 2013 at 18:26
Netro, so you want to just replace text to variable everywhere in the sentence "cat catches dog text"? — SSaikia_JtheRocker
– SSaikia_JtheRocker, Commented Jul 8, 2013 at 18:56

anubhava · Accepted Answer · 2013-07-08 18:41:18Z

2

Use this code:

String s1 = "cat catches dog text";
Pattern p1 = Pattern.compile("\\s*cat\\s+catches\\s*dog\\s+(\\S+)");
Matcher m1 = p1.matcher(s1);
if (m1.find()){
    s1 = m1.replaceFirst(s1.substring(0, m1.start(1)) + "variable");
}
else{
    System.out.println("Else");
}
System.out.println(s1);
// cat catches dog variable

answered Jul 8, 2013 at 18:41

anubhava

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Comments

Ruchira Gayan Ranaweera · Accepted Answer · 2013-07-08 18:29:55Z

1

Try this

import java.util.regex.*;
public class regex1{
public static void main(String args[]){
    String s1 = "cat catches dog text";
    Pattern p1 = Pattern.compile("\\s*cat\\s+catches\\s*dog\\s+(\\S+)");
    Matcher m1 = p1.matcher(s1);
    if (m1.find()){
        System.out.println(m1.group(1));
        s1 = s1.replaceFirst(m1.group(1),"variable");
        System.out.println(s1);
    }
    else{
        System.out.println("Else");
    }
}
}

answered Jul 8, 2013 at 18:29

Ruchira Gayan Ranaweera

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2 Comments

Enigmadan Over a year ago

myMatcher.replaceAll("replacement") has exactly the same results as myString.replaceAll("regex", "replacement"). The difference is just speed. Taken from regular-expressions.info/java.html

Netro Over a year ago

there is subtle diff in using s1.replaceFirst(m1.group(1),"variable"); and m1.replaceFirst(m1.group(1),"variable");. This one solves it easily.

Pshemo · Accepted Answer · 2013-07-08 18:49:45Z

I am not entirely sure what are you trying to do. If you want to replace word after \\s*cat\\s+catches\\s*dog\\s+ with variable then maybe try this way

String s1 = "cat catches dog text";
Pattern p1 = Pattern.compile("(\\s*cat\\s+catches\\s*dog\\s+)(\\S+)");
Matcher m1 = p1.matcher(s1);
if (m1.find()) {
    System.out.println(m1.group(2));
    s1 = m1.replaceFirst("$1variable");
    System.out.println(s1);
} else {
    System.out.println("Else");
}

now group 1 is (\\s*cat\\s+catches\\s*dog\\s+) and you are putting it back to replacement with $1 and adding variable at the end.

output:

text
cat catches dog variable

BTW you don't have to invoke if (m1.find()) if you want to use replaceFirst or replaceAll. Just use it like

String s1 = "cat catches dog text";
Pattern p1 = Pattern.compile("(\\s*cat\\s+catches\\s*dog\\s+)(\\S+)");
Matcher m1 = p1.matcher(s1);
s1 = m1.replaceFirst("$1variable");
System.out.println(s1);

or if you wont need Pattern and Matcher any more just

String s1 = "cat catches dog text";
s1.replaceFirst("(\\s*cat\\s+catches\\s*dog\\s+)(\\S+)","$1variable");

Enigmadan · Accepted Answer · 2013-07-08 19:24:29Z

$1 is a back-referencing method that includes the contents of the 1st ($1) set of parenthesis.

In your case $1 references the first reference created in matcher m1. A reference to the word "text".

Explanation of the code

                        referenced by $1
                             ↓↓↓↓
String s1 = "cat catches dog text";
                                                        reference $1
                                                           ↓    ↓
Pattern p1 = Pattern.compile("\\s*cat\\s+catches\\s*dog\\s+(\\S+)");

Explanation of the regex here. Hover over the colored text to get explanations.

It is important to note that \S (capital 'S') matches any non-whitespace character and the the + is greedy. This means we are getting "all non-whitespace characters before the next whitespace" or simply put, we are getting the next word.

In this case the word being matched happens to be "text"

Matcher m1 = p1.matcher(s1);

m1 will now match "cat catches dog text"

s1 = m1.replaceFirst("variable $1");

s1 is set to s1 ("cat catches dog text") where the first appearance of m1 (the first appearance of '"cat catches dog" followed by any word') is replaced by '"variable" followed by that same word'

If you truly want to have variable replace the word "text" then you need to remove the $1.

s1 = m1.replaceFirst("variable");

s1 is set to s1 ("cat catches dog text") where the first appearance of m1 (the first appearance of '"cat catches dog" followed by any word') is replaced by '"variable"'

I this is the case, you don't actually need to include the parenthesis on the RegEx pattern either. They have no use (in this case) if you are not going to be back-referencing.

Thanks for taking trouble to such great length. This was my thinking also, but this didn't work. It gives ans as variable. Solution given by Ruchira works well.

Mr. Beer · Accepted Answer · 2013-07-10 18:28:46Z

0

String.replaceFirst takes 2 arguments: the regex and the replace string.

So in your example replace,

s1 = s1.replaceFirst("variable $1");

with

s1 = s1.replaceFirst(m1.group(1), "variable");

edited Jul 10, 2013 at 18:28

answered Jul 8, 2013 at 18:24

Mr. Beer

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2 Comments

Enigmadan Over a year ago

myMatcher.replaceAll("replacement") has exactly the same results as myString.replaceAll("regex", "replacement"). The difference is just speed. Taken from regular-expressions.info/java.html

Netro Over a year ago

as suggested by Ruchira, ans is s1 = s1.replaceFirst(m1.group(1), "variable");

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