Creating a random 4 digit number, and storing it to a string

Question

I'm trying to create a method which generates a 4 digit integer and stores it in a string. The 4 digit integer must lie between 1000 and below 10000. Then the value must be stored to PINString. Heres what I have so far. I get the error Cannot invoke toString(String) on the primitive type int. How can I fix it?

   public void generatePIN() 
   {

        //generate a 4 digit integer 1000 <10000
        int randomPIN = (int)(Math.random()*9000)+1000;

        //Store integer in a string
        randomPIN.toString(PINString);

    }

Why do you want it higher than 1000. I think that 0050 is a valid number (though looks strange). — AlikElzin-kilaka, Jun 09 '14 at 10:35
obviously not an exact duplicate but it's the same problem with the same solution...so. — Ryan Haining, Jun 10 '14 at 04:34
For those seeking to generate OTP Pins that could have values like 0004, 0032 or 0516. Here is the code. `Random random = new Random(); String generatePin = String.format("%04d", random.nextInt(10000)); System.out.println(generatePin);` — KNDheeraj, Sep 30 '18 at 16:32
Here is another way. `Random random = new Random(); String generatePin = String.format("%04d", random.nextInt(10000000) % 10000); System.out.println(generatePin);` — KNDheeraj, Sep 30 '18 at 16:33

score 11 · Accepted Answer · answered Mar 06 '14 at 23:34

11

You want to use PINString = String.valueOf(randomPIN);

answered Mar 06 '14 at 23:34

mikejonesguy

9,779
2
35
49

Sharp Edge · Answer 2 · 2014-03-07T00:16:36.763

10

Make a String variable, concat the generated int value in it:

int randomPIN = (int)(Math.random()*9000)+1000;
String val = ""+randomPIN;

OR even more simple

String val = ""+((int)(Math.random()*9000)+1000);

Can't get any more simple than this ;)

edited Mar 07 '14 at 00:16

answered Mar 07 '14 at 00:10

Sharp Edge

4,144
2
27
41

score 2 · Answer 3 · answered Mar 06 '14 at 23:35

2

randomPIN is a primitive datatype.

If you want to store the integer value in a String, use String.valueOf:

String pin = String.valueOf(randomPIN);

answered Mar 06 '14 at 23:35

Alexis C.

91,686
21
171
177

score 1 · Answer 4 · answered Mar 06 '14 at 23:36

1

Use a string to store the value:

   String PINString= String.valueOf(randomPIN);

answered Mar 06 '14 at 23:36

Zied R.

4,964
2
36
67

score 0 · Answer 5 · edited Jan 06 '15 at 15:05

0

Try this approach. The x is just the first digit. It is from 1 to 9.
Then you append it to another number which has at most 3 digits.

public String generatePIN() 
{   
    int x = (int)(Math.random() * 9);
    x = x + 1;
    String randomPIN = (x + "") + ( ((int)(Math.random()*1000)) + "" );
    return randomPIN;
}

edited Jan 06 '15 at 15:05

DoronK

4,837
2
32
37

answered Mar 06 '14 at 23:35

peter.petrov

38,363
16
94
159

What's wrong with `(int)(Math.random()*9000)+1000;` ? – Alexis C. Mar 06 '14 at 23:41
Nothing, I think. I misunderstood what the question was about. I thought it was about the number generation formula itself. – peter.petrov Mar 06 '14 at 23:44

Creating a random 4 digit number, and storing it to a string

5 Answers5