CodeGym/Help with Java Tasks/How regex should looks like in this task?

Level 32

Krakow

15.05.2019
1572views
10comments

How regex should looks like in this task?

Question about the task

Phone number verification

Java Multithreading, Level 2, Lesson 13

Under discussion

How it should be? I added solution for conditions only

The checkPhoneNumber method should check whether the phoneNumber argument is a valid phone number.

Verification criteria:
1) if the number starts with '+', then it must contain 12 digits
2) if the number starts with a digit or opening parenthesis, then it must contain 10 digits
3) the number may contain 0-2 non-consecutive '-' characters
4) the number may contain 1 pair of matching parentheses, which must be to the left of the '-' characters
5) the parentheses must contain exactly 3 digits
6) the number cannot contain letters
7) the number must end with a digit

Here are some examples:

+380501234567 - true
+38(050)1234567 - true
+38050123-45-67 - true
050123-4567 - true
+38)050(1234567 - false
+38(050)1-23-45-6-7 - false
050xxx4567 - false
050123456 - false
(0)501234567 - false

Requirements:

The checkPhoneNumber method should return a boolean.
The checkPhoneNumber method must be public.
The checkPhoneNumber method must have one String parameter.
The checkPhoneNumber method must correctly check the validity of the phone number passed as an argument.

package com.codegym.task.task22.task2212; /* Phone number verification */ public class Solution { public static boolean checkPhoneNumber(String phoneNumber) { String match1 = "\\+\\d{12}"; String match2 = "((\\d{10})|(\\(\\d{3}\\)\\d{7}))"; String match3 = "\\+[(\\d{3}]? "; String match4 = "^\\+\\(\\d{3}\\)\\d{9}"; String match5 = "^\\+\\(\\d{3}\\)\\d{9}"; if (phoneNumber.matches(match1)) return true; if (phoneNumber.matches(match2)) return true; return false; } public static void main(String[] args) { } }

Comments (10)

Popular
New
Old

You must be signed in to leave a comment

Wei Cui

Level 39 , Newark, United States

1 April 2020, 15:59

public static boolean checkPhoneNumber(String phoneNumber) {
        if (phoneNumber == null) return false;

        if (phoneNumber.matches("\\+\\d{12}")) return true;
        //
        if (phoneNumber.matches("^\\+\\d{2}\\(\\d{3}\\)\\d{7}")) return true;

        if (phoneNumber.matches("^\\+\\d{8}-\\d{2}-\\d{2}")) return true;

        if (phoneNumber.matches("^\\d{6}-\\d{4}")){return true;}
        return false;
    }

BlueJavaBanana

Level 37

15 July 2020, 15:55

This looks like you simply wrote a regex fo each of the example phone numbers in the task? Is this so? What would happen for instance if you gave your method "+-(000)123456789

null

Level 26 , Orlando, United States

4 August 2020, 02:13

I feel stupid after reading this piece of code. lol

Artur Jarosz

Level 30 , Lublin, Poland

9 February 2020, 21:31

First check requirements regarding dashes. Then you can remove them and then one regex is possible. :)

Guadalupe Gagnon

Level 37 , Tampa, United States

17 May 2019, 14:15

I solved this by first determining the length of the number when you removed everything that wasn't a digit. Here is an outline of my solve:

String originalNum = "+38050123-45-67";
int length = originalNum.replaceAll("[^0-9]","").length();

if(length == 12){
            // starts with '+'
            // can have 0 or a few '#'
            // can have 1 or none '(###)'
            //can have 1 or none "-#"
            //can have 1 or none "#-#"
           // return true if match
} else if(length == 10){
            // check for 10 digit number
           // return true if match
}

return false;

I left this fairly abstract, but it is a description of how I solved. You may be able to find a better way, maybe just 1 regex that works for any number according to task conditions.

Johannes

Level 27 , Centurion, Pretoria, South-Africa

29 April 2020, 13:08

Guada: how does your replaceAll work ? I know it replace all occurrences of param1 with param2. e.g. replaceAll("a","b") will turn "abac" into "bbbc". So how come your first parameter (all numbers) is not replaced with "", but instead all the NON-numbers are replaced with "" ? I don't get it ?

Johannes

Level 27 , Centurion, Pretoria, South-Africa

29 April 2020, 13:12

Also, i tried using "^\\+" as regular expression, and it doesn't work ? (if the string starts with a plus sign) if (phoneNumber.matches("^\\+"))

Guadalupe Gagnon

Level 37 , Tampa, United States

29 April 2020, 14:04

It is because, inside the brackets [], the ^ symbol means NOT. So the first parameter reads "replace everything that is not a digit between 1-9". "^\\+" would only return true for a string that was just "+", if you wanted to check that a String such as "+##(###)##-####" starts with "+" using regex you would need to tell it, after the "^\\+" that it can contain any amount of characters following. So the correct regex for this would be "^\\+.*". You could also simply use: if(string.charAt(0) == '+', which may be simpler than figuring out a proper regex. There is a saying that goes - if you choose to solve a problem with a regex then you will have 2 problems.

Guadalupe Gagnon

Level 37 , Tampa, United States

29 April 2020, 14:10

Regular expressions are not an easy subject and could have a whole course dedicated for them. They are basically a whole different programming language. It takes a lot of practice to start to get a grasp on them, and I'm afraid I only know the very basics. I do like to use this website as a reference: https://www.rexegg.com/regex-quickstart.html

Johannes

Level 27 , Centurion, Pretoria, South-Africa

30 April 2020, 01:10

ha!ha! Thanks man! I wanted to use normal string functions like you suggest, but thought the validator WANTED regex's. Thanks, makes sense now.