Check if a string match a defined pattern

Question

I am working on an assignment right now where I have to enter an input in "ABC123456" format entry and see if it is valid or not. I don't understand how to check each character if it is a number or letter. Here is what I have so far:

import java.util.Scanner;
public class NetID {


    public static void main(String[] args) {    
        Scanner input = new Scanner (System.in);
        String userInput, char0thru2, char3thru8, uppercase, netID;

        System.out.println("Please enter the NetID to verify:");
        userInput = input.nextLine();

        if (userInput.length() != 9) {
            System.out.println("Your NetID needs to be 9 characters long, it needs to be in this format: ABC123456");
        }   
        if (userInput.length() == 9){
            char0thru2 = userInput.substring(0, 3);
            char3thru8 = userInput.substring(3, 9);
            uppercase = char0thru2.toUpperCase();
        }
    }
}

https://docs.oracle.com/javase/8/docs/api/java/lang/Character.html#isLetter-char-, https://docs.oracle.com/javase/8/docs/api/java/lang/Character.html#isDigit-char- — JB Nizet, Oct 22 '17 at 10:56
https://docs.oracle.com/javase/8/docs/api/java/util/regex/Pattern.html — M. le Rutte, Oct 22 '17 at 10:56

score 1 · Accepted Answer · answered Oct 22 '17 at 11:26

Just use a pattern:

String userInput = "ABC133456";

    if(!Pattern.matches("[A-Z]{3}[0-9]{6}", userInput))
         System.out.println("Your NetID needs to be 9 characters long, it needs to be in this format: ABC123456");
    else
         System.out.println("Ok!");

Nutan · Answer 2 · 2017-10-22T11:36:38.683

0

You can try this way

Loop over the input string, and get each character at the counter position of the loop , then compare the ASCCI values of each character. If u want first 3 character as letters then till the counter is 2 compare them with ASCCI value of A-Z and all other with ASCCI value of 0-9

You can try below code , may have some compilation errors as I haven't compiled it :(

  if(userInput.length==9)
   {
     for (int i =0;i<userInput.lenght;i ++){
       if (i <3){
        if(!(userInput.charAt(i)<91      &&userInput.charAt(i)>64)){
         System.out.println("invalid string");
          break;
        }
      }
      if (i >2){
      if(!(userInput.charAt(i)<58 &&userInput.charAt(i)>47)){
       System.out.println("invalid string");
        break;
      }
     }
    }
   }

edited Oct 22 '17 at 11:36

answered Oct 22 '17 at 11:05

Nutan

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I have only recently started to learn coding and I am not really sure I understand what you mean – Zain Kalwani Oct 22 '17 at 11:09
U have a constant format to get string ? i.e first 3 letters always be letter and last 3 always be number ? And string will be of length 9 ? – Nutan Oct 22 '17 at 11:15
Yes the first three are letters and last 6 are numbers and length is 9 – Zain Kalwani Oct 22 '17 at 11:16
Added a code ,please try to remove compilation errors if any and try to execute it will work – Nutan Oct 22 '17 at 11:37

score 0 · Answer 3 · answered Oct 22 '17 at 11:19

Simply iterate over String and use isDigit() of Character class,sample code below, hope it helps:

String inp="qwerty43";
for(int i=0;i< myStr.length();i++){
       if (Character.isDigit(myStr.charAt(i))) {
            System.out.println("Digit Found");
        } else {
            System.out.println("Letter Found");
        }
}

Check if a string match a defined pattern

3 Answers3