Java - differentiating between strings

Question

Is there a number wildcard character in java? I'm opening a file and looking at a list of data and I need to differentiate between three pieces of information that start with "M". However, one of them has numbers directly following it and the other two have letters that follow. I was wondering if there was a way to check if there was a number after the letter with a wildcard character. I'm sure you could do this with ASCII, but I also am unsure of how to execute that.

EDIT: I'm still having issues, so here is my code.

import java.io.*;
import java.util.*;
import java.util.regex.*;

public class addSevTest{
    public static void main(String[] args) throws IOException{
        FileReader fr = new FileReader("output6.txt");
        BufferedReader br = new BufferedReader(fr);
        String line;
        Pattern pattern = Pattern.compile(br.readLine());
        Matcher matcher = pattern.matcher(br.readLine());

    List<String> list = new ArrayList<String>();
    while ((line = br.readLine()) != null){
        if(line.contains("100%") || line.contains("70%") || matcher.find("[.][1-9]")){
            list.add(line);
            list.add("     2");
            list.add("\n");
            //System.out.println('Using String matches method: '+line.matches('.M'));
        }else if(line.startsWith("MDRALM")){
            list.add(line);
            list.add("     3");
            list.add("\n");
        }else if(line.startsWith("SOL") || line.startsWith("I/O") || line.startsWith("AH") || line.startsWith("LT")){
            continue;
        }else{
            list.add(line);
            list.add("     1");
            list.add("\n");
        }
    }

    /*while ((line = br.readLine()) != null){
        if(line.contains("CP")){
            list.add(line);
            list.add("\n");
        }
    }*/
    br.close();

    FileWriter writer = new FileWriter("addSevTest_O.txt"); 
        for(String str: list){
            writer.write(str);
        }
        writer.close();
}

}

Regular expressions. See the API documentation of the package java.util.regex. — Jesper
– Jesper, Commented Sep 5, 2013 at 14:42
You're looking for a Regex. You can use Pattern and Matcher classes along with your Regex to evaluate the String. — Luiggi Mendoza
– Luiggi Mendoza, Commented Sep 5, 2013 at 14:42
Regex is a language on it's own. Whether you're programming ruby, javascript, java, or even C.. Regex, while will have little differences between languages... it is at it's core, a language in and of itself. Have fun with it. — ddavison
– ddavison, Commented Sep 5, 2013 at 14:57

StormeHawke · Accepted Answer · 2013-09-05 17:42:25Z

5

You'd be best off using some simple regular expressions.
I found some basic tutorials you can skim through for the basics here:

And a couple of tools to help you on your journey:

EDIT

In your added code, try replacing this:

    if(line.contains("100%") || line.contains("70%") || matcher.find("[.][1-9]"))

with this:

    if(line.contains("100%") || line.contains("70%") || line.matches("M[1-9]+.*"))

The M matches the first letter of the line. [1-9] matches the digits, with the + meaning one or more. .* means zero or more additional characters following the number will also match.

The Pattern/Matcher stuff you've got here is overkill for your purposes.

edited Sep 5, 2013 at 17:42

answered Sep 5, 2013 at 14:43

StormeHawke

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7 Comments

StormeHawke Over a year ago

@CinCity add the code as an edit to your original question and we'll look

StormeHawke Over a year ago

@CinCity To start, pattern.compile should take a regular expression string.

StormeHawke Over a year ago

Secondly, using Pattern and Matcher might be overkill here... a simple string.matches(regex) would probably serve your purposes - something like string.toLowerCase().matches("M[0-9].*") might do the trick

StormeHawke Over a year ago

Oh, and this regex - "[.][1-9]" what are you trying to match with [.]? Because that's going to match the literal character '.'

CinCity Over a year ago

Totally works, many thanks my good sir. I'm going to have to study the regex harder.

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