Wrap the string after a number of characters word-wise in Java

Question

I have this code:

    String s = "A very long string containing " +
                   "many many words and characters. " +
                   "Newlines will be entered at spaces.";

    StringBuilder sb = new StringBuilder(s);

    int i = 0;
    while ((i = sb.indexOf(" ", i + 20)) != -1) {
        sb.replace(i, i + 1, "\n");
    }

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

The output of the code is:

A very long string containing
many many words and
characters. Newlines
will be entered at spaces.

The above code is wrapping the string after the next space of every 30 characters, but I need to wrap the string after the previous space of every 30 characters, like for the first line it will be:

A very long string

And the 2nd line will be

containing many

Please give some proper solution.

Answer 1

You can use Apache-common's WordUtils.wrap().

Answer 2

Use lastIndexOf instead of indexOf, e.g.

StringBuilder sb = new StringBuilder(s);

int i = 0;
while (i + 20 < sb.length() && (i = sb.lastIndexOf(" ", i + 20)) != -1) {
    sb.replace(i, i + 1, "\n");
}

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

This will produce the following output:

A very long string
containing many
many words and
characters.
Newlines will be
entered at spaces.

Answer 3

You can try the following:

public static String wrapString(String s, String deliminator, int length) {
    String result = "";
    int lastdelimPos = 0;
    for (String token : s.split(" ", -1)) {
        if (result.length() - lastdelimPos + token.length() > length) {
            result = result + deliminator + token;
            lastdelimPos = result.length() + 1;
        }
        else {
            result += (result.isEmpty() ? "" : " ") + token;
        }
    }
    return result;
}

call as wrapString("asd xyz afz","\n",5)