Question

I was trying out to create a regular expression to match file path in java like

C:\abc\def\ghi\abc.txt

I tried this ([a-zA-Z]:)?(\\[a-zA-Z0-9_-]+)+\\? , like following code

import java.util.regex.Pattern;

  public class RETester {

public static void main(String arhs[]){

    String regularExpression = "([a-zA-Z]:)?(\\[a-zA-Z0-9_-]+)+\\?";

    String path = "D:\\directoryname\\testing\\abc.txt";

    Pattern pattern = Pattern.compile(regularExpression);

    boolean isMatched = Pattern.matches(regularExpression,path);
    System.out.println(path);
    System.out.println(pattern.pattern());
    System.out.println(isMatched);

}

}

However it’s always giving me , false as result .

Answer 1

Java is using backslash-escaping too, you know, so you need to escape your backslashes twice, once for the Java string, and once for the regexp.

"([a-zA-Z]:)?(\\\\[a-zA-Z0-9_.-]+)+\\\\?"

Your regexp matched a literal ‘[-zA-Z0-9_-‘ string, and a literal ‘?’ at the end. I also added a period in there to allow ‘abc.txt’..

That said, consider using another mechanism for determine valid file names, as there are different schemes (i.e. unix). java.util.File will probably throw an exception if the path is invalid, which might be a good alternative, although I don’t like using exceptions for control flow…