how to generate sequence random number in c#?

Question

Hi Can Any One tell me How to generate sequence random number in C# Windows form.... my expected answer is: 00 01 02 03

Vulpes · Answer

OK in that case this should work: private void button1_Click(object sender EventArgs e) { for (int i = 0; i { comboBox1.Items.Add(i.ToString(D2)); } }

Vulpes · Answer

The D2 format takes an integer and converts it to a string (at least) 2 digits long. If it's an integer between 0 and 9 it's prefixed with an 0. In a simple case such as this you could also do: for (int i = 0; i { comboBox1.Items.Add(0 + i.ToString()); }

selva kumar · Answer

what is the purpose of using D2?

selva kumar · Answer

yes i want the out put like this... 00 01 02 03

Vulpes · Answer

Do you mean that the first number should be random but the next 3 should then be sequential? For example: 23 24 25 26 or do you mean that the numbers should always be: 00 01 02 03

selva kumar · Answer

hi It's working....but it doesn't have a sequence order..... like this... 00 01 02 03 can u tell me another way....

Vulpes · Answer

To put them in a combobox you can use: private void button1_Click(object sender EventArgs e) { for (int i = 0; i { int number = rand.Next(100); // random number between 0 and 99 comboBox1.Items.Add(number.ToString(D2)); } } The numbers you gave as an example are not random they're just sequential. Mine are random two digit numbers between 00 and 99. Please clarify what it is you need.

selva kumar · Answer

i want to display the out put in dropdown list ......ur answer does't match for my exact answer... can u tell me another way......

Vulpes · Answer

Try this: private static Random rand = new Random(); private void button1_Click(object sender EventArgs e) { string message = ; for (int i = 0; i { int number = rand.Next(100); // random number between 0 and 99 message += number.ToString(D2) + 
; } MessageBox.Show(message); }