Need some simple help on using .split in java

Question

I have a simple code:

    public static void main(String[] args) {
        String s = "He is a very very good boy, isn't he?"
        String[] words = s.split("[\\s\\-\\.\\'\\?\\,\\_\\@\\!]");
        System.out.println(words.length);
        for(int i = 0; i<words.length; i++) {
            System.out.println(words[i]);
        }
        scan.close();
    }

that should output me this:

10
He
is
a
very
very
good
boy
isn
t
he

But instead, it prints me out this:

11
He
is
a
very
very
good
boy

isn
t
he

Can anyone suggest me how to fix this issue? I know that the problem is when my program encounter "," it automatically splits the string and then again followed by " " it splits it again so it creates a empty line in my output, but i have no idea how to fix it so it will split multiple delimiters at the same time.

Sweeper · Accepted Answer · 2020-05-06 14:50:36Z

2

First, although characters with special meaning in regex like ? and . in general has to be escaped, they don't need to be escaped when they are in a character class, [].

So your split call is equivalent to:

String[] words = s.split("[\\s\\-.'?,_@!]");

Only - needs to be escaped because it means "to" in a character class.

Essentially you want it to treat ", " as one delimiter. To match one or more characters, you should use the + quantifier:

String[] words = s.split("[\\s\\-.'?,_@!]+");

Here you are saying that a delimiter is at least one of those characters in the character class.

Here are some visualisations to see what characters are matched, compare:

with +:
without +:

edited May 6, 2020 at 14:50

answered May 6, 2020 at 14:44

Sweeper

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6 Comments

JD_KX Over a year ago

Okay thank you it works as it should now :D But what does escaped mean?

Sweeper Over a year ago

In your regex, you have used lots of backslashes haven't you? Why? @JD_KX

JD_KX Over a year ago

No idea what i used them for to be honest, I just seen somewhere in other people stackoverflow posts that they put them and it works so i did :P

Sweeper Over a year ago

@JD_KX Okay then. Backlashes escape the character after them. This means that whatever is after a backslash loses its original meaning. For example, . originally means "any non-new-line character", but if you write a backslash before it, it means "the dot character". However, in character classes, many characters with special meanings are automatically escaped, so you don't need backslashes. Many people put backlashes in anyway (as a habitual thing?).

JD_KX Over a year ago

r u there? Hey can u help me with thing that makes my program work a bit not as expected? To be exact, when i assign to my string a value "" (string is empty but it does allocate a value in array), my system.out.println(words.length) returns 1 but i want it to return 0 because this program does not split anything, how do i fix that?

|

WJS · Accepted Answer · 2020-05-06 14:47:57Z

1

Try it this way.

Replace all the characters you don't want with spaces
Then split on one or more spaces.

String s = "He is a very very good boy, isn't he?";
String[] words = s.replaceAll("[\\W]"," ").split("\\s+");
System.out.println(words.length);
for(int i = 0; i<words.length; i++) {
   System.out.println(words[i]);
}

Or just use split on non words

String[] words = s.split("\\W+");

answered May 6, 2020 at 14:47

WJS

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Comments

Albert Turri · Accepted Answer · 2020-05-06 14:48:14Z

1

public static void main(String[] args) {
    String s = "He is a very very good boy, isn't he?";
    String[] words = s.split("([\\s\\-.\\'\\?\\,\\_\\@\\!])+");
    System.out.println(words.length);
    for (String word : words) {
        System.out.println(word);
    }
}

answered May 6, 2020 at 14:48

Albert Turri

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