How to generate 6 digit random number - dart

Question

I want to generate a random six-digit number. I tried to use the Random class, but new Random().nextInt(999999) generates some numbers with less than six digits.

Answer 1

So you want just the numbers 100000 to (and including) 999999.
you can get a random number in this range (900000) and add 100000 to the random number you get:
var rng = new Random();
var code = rng.nextInt(900000) + 100000;
This will always give you a random number with 6 digits.

Answer 2

void main() {
var rnd = new math.Random();
var next = rnd.nextDouble() * 1000000;
while (next < 100000) {
next *= 10;
}
print(next.toInt());
}

Answer 3

you can also generate 6 different numbers and then concatenate them in one string and convert it to integer if you want
import 'dart:math';
main(){
var rndnumber="";
var rnd= new Random();
for (var i = 0; i < 6; i++) {
rndnumber = rndnumber + rnd.nextInt(9).toString();
}
print(rndnumber);
}

Answer 4

Here is a Dart extension method that will generate a non-negative random integer with a specified number of digits:
extension RandomOfDigits on Random {
/// Generates a non-negative random integer with a specified number of digits.
///
/// Supports [digitCount] values between 1 and 9 inclusive.
int nextIntOfDigits(int digitCount) {
assert(1 <= digitCount && digitCount <= 9);
int min = digitCount == 1 ? 0 : pow(10, digitCount - 1);
int max = pow(10, digitCount);
return min + nextInt(max - min);
}
}
In your case use it like this:
final random = Random();
print(random.nextIntOfDigits(6));

Answer 5

The following class will generate an integer with 'n' digits, or a string with 'n' digits.
The numeric method will be much faster, but is limited in the number of digits.
import 'dart:math';
class RandomDigits {
static const MaxNumericDigits = 17;
static final _random = Random();
static int getInteger(int digitCount) {
if (digitCount > MaxNumericDigits || digitCount < 1) throw new RangeError.range(0, 1, MaxNumericDigits, "Digit Count");
var digit = _random.nextInt(9) + 1; // first digit must not be a zero
int n = digit;
for (var i = 0; i < digitCount - 1; i++) {
digit = _random.nextInt(10);
n *= 10;
n += digit;
}
return n;
}
static String getString(int digitCount) {
String s = "";
for (var i = 0; i < digitCount; i++) {
s += _random.nextInt(10).toString();
}
return s;
}
}
void main() {
print(RandomDigits.getInteger(6));
print(RandomDigits.getString(36));
}
Output:
995723
198815207332880163668637448423456900

Answer 6

If you want to get 6 digit value from 0 to 999999, you can add leading 0 if the number is less than 6 digits.
String r = Random().nextInt(999999).toString().padLeft(6, '0');
// example output: 025328

Answer 7

i had the same problem.although there are 3-4 ways to tackle it but i find the below one simple and sorted.
simply check for the number of length match. find below code
Integer otp = new Random().nextInt(999999);
int noOfOtpDigit=6;
while(Integer.toString(otp).length()!=noOfOtpDigit) {
otp = new Random().nextInt(999999);
}
String otpString = String.valueOf(otp);

Answer 8

import 'dart:math';
void main() {
print(get6DigitNumber());
}
String get6DigitNumber(){
Random random = Random();
String number = '';
for(int i = 0; i < 6; i++){
number = number + random.nextInt(9).toString();
}
return number;
}

Answer 9

Here is your unlimited supply of six-digit random numbers
String getRandomNumber(){
final r = Random();
return List<int>.generate(6, (index) => r.nextInt(10)).fold<String>("", (prev, i) => prev += i.toString());
}

Answer 10

This is working for me in C#.
Random random = new Random();
string elementIndex = random.Next(100000, 999999).ToString();