How to delete the ends of line ["\n"] in a String before parsing

Question

hi, i'd want to parse a String with Regex/Pattern.
This String contains ends of line i want to delete before applying my parsing treatment.
I've tried to do that by replacing the character "\n" with "" :
    // the original text
    String orgTextToParse = new String(buffer);
    //we create a pattern to recognize the ends of line
    String regExpEol = "\n";
    Pattern patternEol = Pattern.compile(regExpEol);
    Matcher matcherEol = patternEol.matcher(orgTextToParse);
    // the new text without end of line
    StringBuffer newTextToParse = new StringBuffer();
    // we replace the "\n" by ""
    while(matcherEol.find()) {
      String REPLACE = "";
      matcherEol.appendReplacement(newTextToParse,REPLACE);
    matcherEol.appendTail(newTextToParse);
    System.out.println("[new string start] " + (newTextToParse.toString()) + " [end of new string]");Is there an appropriate way to delete "\n" ?
(perhaps a method of the Pattern or Matcher classes... ?)
thanx
Edited by: jfact on Dec 29, 2007 11:05 AM
Edited by: jfact on Dec 29, 2007 11:06 AM

Answer 1

The second (optional) argument to Pattern is a set of flags. Take a look at the Javadoc.
The 'genomic' implication is that the EOL is an arbitrary line break for human consumption so it does look like you need to remove the EOL. I would just use replaceAll() since it can be done with one line of code (or just a couple if you pre-compile the regex) without the need for any looping.
line = line.replaceAll("\n","");Edited by: sabre150 on Dec 29, 2007 10:27 AM