Regex validation goes wrong - Java

Question

So I have this row of regex to check my input. There is an input because the else is activated but I can't find it.

Username: Donald
Password: 1234
Name: Donald
LastName: Muylle
Phone: 012121212
Address: Zeverstreet 12
City: Gestel
Job: Bar

if(txtUsername.getText().matches("[a-zA-Z0-9]")
&& txtPassword.getText().matches("[a-zA-Z0-9]") &&
txtName.getText().matches("[a-zA-Z]") && 
txtLastName.getText().matches("[a-zA-Z]") && 
txtPhone.getText().matches("[0-9]") &&
txtAddress.getText().matches("[a-zA-Z0-9\\s]") && 
txtCity.getText().matches("[a-zA-Z]") && 
txtJob.getText().contains("[a-zA-Z]")){
    // Code
} else {
    // Shows Alert
}

Avinash Raj · Accepted Answer · 2015-03-07 10:30:35Z

1

All you have to do is to add + quantifier next to all the character classes like,

txtUsername.getText().matches("[a-zA-Z0-9]+")
                                          ^

So that, it would match one or more occurrences of the characters given in the list. Without +, it would match a single character only where the matches method must check for a match against the whole string. So this would end up in failure if the input contain more than one character.

answered Mar 7, 2015 at 10:30

Avinash Raj

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5 Comments

Benjamin Naesen Over a year ago

For some weird reason, still showing my error. It didn't fix the problem. Thanks though.

Benjamin Naesen Over a year ago

It's an error button I made my self that just says 'Credentials not filled in correctly.'

Avinash Raj Over a year ago

it seems like error caused by some part of your program not the regex.

Benjamin Naesen Over a year ago

I already debugged all the rest and ended up in the regex because if I system.out.println('bla') in the ELSE part, it would show up, so that makes my attention go to the regex being a failure.

Benjamin Naesen Over a year ago

I think I found it, the last matches is 'contains' and not 'matches'. How did I get that :D Still thanks for the help!

Wiktor Stribiżew · Accepted Answer · 2015-03-07 10:42:27Z

1

You have contains in the last check. Replace it with matches.

answered Mar 7, 2015 at 10:42

Wiktor Stribiżew

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Comments

Davide Lorenzo MARINO · Accepted Answer · 2015-03-07 10:47:19Z

1

All your regular expressions match only strings of 1 char. Modify all them adding a + symbol at the end.

So [a-zA-Z0-9] becomes [a-zA-Z0-9]+, [a-zA-Z] becomes [a-zA-Z]+ and so on.

The address regular expression [a-zA-Z0-9\\s] should be [a-zA-Z0-9\s]+ with only one slash

edited Mar 7, 2015 at 10:47

answered Mar 7, 2015 at 10:35

Davide Lorenzo MARINO

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3 Comments

Benjamin Naesen Over a year ago

For some weird reason, still showing my error. It didn't fix the problem. Thanks though.

Davide Lorenzo MARINO Over a year ago

There is also another problem in the address regular expression.

Davide Lorenzo MARINO Over a year ago

There is also another problem in the address regular expression.

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