Using StringBuildersetLength() instead of StringBuilder. deleteCharAt() when we remove trailing zeroes because it also deletes the last few characters and it's more performant.
Regex with replaceAll.
public class Main
{
    public static void main(final String[] argv) 
    {
        String str;
        str = "hello\r\njava\r\nbook";
        str = str.replaceAll("(\\r|\\n)", "");
        System.out.println(str);
    }
}
If you only want to remove \r\n when they are pairs (the above code removes either \r or \n) do this instead:
str = str.replaceAll("\\r\\n", "");
    If you want to avoid the regex, or must target an earlier JVM, String.replace() will do:
str=str.replace("\r","").replace("\n","");
And to remove a CRLF pair:
str=str.replace("\r\n","");
The latter is more efficient than building a regex to do the same thing. But I think the former will be faster as a regex since the string is only parsed once.
public static void main(final String[] argv) 
{
    String str;
    str = "hello\r\n\tjava\r\nbook";
    str = str.replaceAll("(\\r|\\n|\\t)", "");
    System.out.println(str);
}
It would be useful to add the tabulation in regex too.
Given a String str:
str = str.replaceAll("\\\\r","")
str = str.replaceAll("\\\\n","")
    
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