BufferedReader won't read

Question

I have a program that the username and password are in a text file and the text file looks like this:

election:12345

and I have this code to read the file

try {
    BufferedReader read=new BufferedReader(new FileReader("election_un_pass.txt"));
    String line="";

    while((line=read.readLine())!=null) {
        String [] info=line.split(":");

        if(info[0].matches(Login.uname) && info[1].matches(Login.pass)){
            new Main();
        } else {
            JOptionPane.showMessageDialog(null, "Username or Password might not be correct");
        }
        Login.txtUName.setText("");
        Login.txtPassword.setText("");
     }
} catch (Exception e1) {
    e1.printStackTrace();
}

Every time I run my program, even tho the username and password that I entered are correct, the Username or Password might not be correct message will still appear and new Main() won't appear.

Please check the Login.uname and Login.pass values it seems they are not matching. — Vural, Sep 07 '16 at 21:38
Sure that you don't want to use `String.equals`? And what are the values of `Login`? — Marvin, Sep 07 '16 at 21:40
@VuralAcar oh they are actually matched and the Login.uname and Login.pass are initialized to the txtUName.getText() and txtPassword.getText() but because of your suggestion i just directly used the txtUName.getText() and txtPassword.getText(). thank you so much!!!! — luh, Sep 07 '16 at 21:45
@Marvin `txtUName.getText()` and `txtPassword.getText()` but i just directly used them now and it's now fixed thank you so much!!! — luh, Sep 07 '16 at 21:47
Possible duplicate of [What is a debugger and how can it help me diagnose problems?](http://stackoverflow.com/questions/25385173/what-is-a-debugger-and-how-can-it-help-me-diagnose-problems) — Raedwald, Sep 11 '16 at 08:12

score 0 · Accepted Answer · answered Sep 07 '16 at 21:50

You could:

Learn how to debug in whatever IDE you are using. With debug mode you can execute the code one line at a time and see the values of the variables at each point.
Increase the logging information when the error occurrs

For example you could print to the console/logs/dialog:

String msg = String.format("could not find user name: [%s] and password: [%s] in line: [%s]", Login.uname, Login.pass, line);

Obviously you shouldn't print the expected user name and password anywhere when an incorrect value is input for security reasons, but it might be a quick way to find out whats going wrong :) It could be as simple as trailing whitespace characters somewhere or incorrect case somewhere.

score 0 · Answer 2 · answered Sep 07 '16 at 22:11

Are you certain that you want to use String.matches(regex) for your comparisons? The argument is taken as a regular expression, so any regex meta-characters in your Login.uname or Login.pass values would be interpreted as regex. This means your match might work in ways you don't expect.

I suspect you really want to say:

if (info[0].equals(Login.uname) && info[1].equals(Login.pass)) {

instead.

BufferedReader won't read

2 Answers2