Generate random string into text box (code inside)

Question

1.00/5 (1 vote)

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Hi guys, I need help with my queue number generator.

Basically, i need to generate a 5 character string with the letter 'A' at the front followed by 4 random numbers

example:

A6932
A1038
A0193

C#

Random random = new Random();
int length = 5;

for (int i = 0; i < length; i++)
{
	//if random.Next() == 0 then we generate a random character
	if (random.Next(0, 3) == 0)
	{
		tb_queuenum.Text += ((char)random.Next(65, 91)).ToString();
	}
	else //if random.Next() == 0 then we generate a random digit
	{
		tb_queuenum.Text += random.Next(0, 9);
	}
}

These are my codes right now, any help will be deeply appreciated <3
If there's a better alternative to my codes please share.
God bless.

Posted 6-Jul-15 10:49am

Floyd Pinkerton

Updated 6-Jul-15 10:59am

virusstorm

v2

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[no name] 6-Jul-15 16:58pm

Help... with what? You did not describe any kind of problem....

1 solution

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virusstorm · Accepted Answer · 2015-07-06T11:07:00

Solution 1

Not sure what exactly your issue is, but you did ask if there was a better way to achieve this.

C#

int min = 0;
int max = 9999;

Random random = new Random();
int randomInt = random.Next(min, max);

var randomValue = String.Format("A{0:D4}", randomInt);

Posted 6-Jul-15 11:07am

virusstorm