Java Regex Metacharacters returning extra space while spliting

Question

I want to split string using regex instead of StringTokenizer. I am using String.split(regex); Regex contains meta characters and when i am using \[ it is returning extra space in returning array.

import java.util.Scanner;
public class Solution{
    public static void main(String[] args) {
        Scanner i= new Scanner(System.in);
        String s= i.nextLine();
        String[] st=s.split("[!\\[,?\\._'@\\+\\]\\s\\\\]+");
        System.out.println(st.length);
        for(String z:st)
            System.out.println(z);
        }
}

When i enter input [a\m] It returns array length as 3 and

a m

Space is also there before a. Can anyone please explain why this is happening and how can i correct it. I don't want extra space in resulting array.

You're just getting an empty string as first element in array because your input starts with `[` and splitting is done on `[` as well. — anubhava, Jan 20 '16 at 18:24
Remove all these characters from the start. http://ideone.com/PXWLYg — Wiktor Stribiżew, Jan 20 '16 at 18:26
@WiktorStribiżew Wow Thanks. Great Code but i have query in your code. If it is replacing all these then how it is splitting the string? — Linkon, Jan 20 '16 at 18:30
No, it does not remove *all*, only those at the start. You are not interested in those leading symbols since they produce empty array elements, right? So, removing them at the start of string only is a valid approach. — Wiktor Stribiżew, Jan 20 '16 at 18:36
Yes, this is exactly what i want. Can you explain little bit about it's working if you got 1 or 2 minutes free? replaceAll("^[!\\[,?._'@+\\]\\s\\\\]+", "") like ^ negates then how it is replacing? I am new to this :( — Linkon, Jan 20 '16 at 18:39

Wiktor Stribiżew · Accepted Answer · 2016-01-20T18:40:37.717

3

Since the [ is at the beginning of the string, when split removes [, there appear two elements after the first split step: the empty string that is at the beginning of the string, and the rest of the string. String#split does not return trailing empty elements only (as it is executed with limit=0 by default).

Remove the characters you split against from the start (using a .replaceAll("^[!\\[,?._'@+\\]\\s\\\\]+", note the ^ at the beginning of the pattern). Here is a sample code you can leverage:

String[] st="[a\\m]".replaceAll("^[!\\[,?._'@+\\]\\s\\\\]+", "")
                 .split("[!\\[,?._'@+\\]\\s\\\\]+");
System.out.println(st.length);
for(String z:st) {
    System.out.println(z);
}

See demo

edited Jan 20 '16 at 18:40

answered Jan 20 '16 at 18:28

Wiktor Stribiżew

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Thanks a lot for your detailed Explanation and Code :) – Linkon Jan 20 '16 at 18:42
Sir, if i input only ! or [ it is still returning 1 whereas it should return 0 :( – Linkon Jan 20 '16 at 19:12
2

It is because when you split an empty string, you will get 1 empty element. Use [this code](http://ideone.com/4iHNHK) to handle that situation. – Wiktor Stribiżew Jan 20 '16 at 19:24
2

Since Java 6, you can use `string.isEmpty()` instead of `string.length() != 0`, but that’s only a stylistic difference… – Holger Jan 20 '16 at 19:27
I did it using "\\\\" instead of "" while replacing Now it returns 0 – Linkon Jan 20 '16 at 19:28
String[] st=s.replaceAll("^[!\\[,?._'@+\\] \\\\]+","\\\\").split("[!\\[,?\\._'@\\+\\]\\s\\\\]+"); – Linkon Jan 20 '16 at 19:28

score 2 · Answer 2 · edited May 23 '17 at 12:15

As an addition to Wiktor Stribiżew’s answer, you may do the same without having to specify the pattern twice, by dealing with the java.util.regex package directly. Removing this redundancy may avoid potential errors and may also be more efficient as the pattern doesn’t need to be parsed twice:

Pattern p = Pattern.compile("[!\\[,?\\._'@\\+\\]\\s\\\\]+");
Matcher m = p.matcher(s);
if(m.lookingAt()) s=m.replaceFirst("");
String[] st = p.split(s);
for(String z:st)
    System.out.println(z);

To be able to use the same pattern, i.e. without having to use the anchor ^ for removing a leading separator, we first check via lookingAt() whether the pattern really matches at the beginning of the text before removing the first occurrence. Then, we proceed with the split operation, but reusing the already prepared Pattern.

Regarding your issue mentioned in a comment, the split operation will always return at least one element, the input string, when there is no match, even when the string is empty. If you wish to have an empty array then, the only solution is to replace the result explicitly:

if(st.length==1 && s.equals[0]) st=new String[0];

or, if you only want to treat an empty string specially, you may check this beforehand:

if(s.isEmpty()) st=new String[0];
else {
  // the code as shown above
}

Java Regex Metacharacters returning extra space while spliting

2 Answers2