I want to generate a random number sequence but ...

Question

I want to generate a random number sequence but .. I want to be able to change sequence numbers count .

Also I don't need to show something I tryed to find a way to get over comboBox but I failed :(

Any advise ?? :)

I found something like this it doesn't work :(( ;

private void button1_Click(object sender, EventArgs e)
{
for (int i = 0; i < 4; i++)
{
comboBox1.Items.Add(i.ToString("D2"));
}
}

Amit Gupta · Answer

Try like this System.Random rand = new System.Random(); for (int i = 0; i < 4; i++) { int number = rand.Next(180); comboBox1.Items.Add(number.ToString("D2")); }

Ravi Patel · Answer

Hi, there is no next() such method it will be in capital letter Next() its not rand.next(180) it will be rand.Next(180); private void button1_Click( object sender, EventArgs e) { for ( int i = 0; i { System.Random rand = new System.Random(); int number = rand.Next(180); comboBox1.Items.Add(number.ToString( "D2" )); } }

Laxmidhar Sahoo · Answer

hi public static void genaraterandom() { var _random = new Random(); for ( int i = 0; i { var r = _random.Next(100, 1000); } } Thanks and regards

Amit Gupta · Answer

In c#, method name always starts with upper case, so Next(..) should starts with upper case

Ömer Faruk Çalışkan · Answer

I am trying but this thing says ; Assets\Lock.cs(22,35): error CS1061: 'Random' does not contain a definition for 'next' and no accessible extension method 'next' accepting a first argument of type 'Random' could be found (are you missing a using directive or an assembly reference?) and also I need to get number by writeline.. comboBox don't do for me :D sorry I posted a wrong code script this is what I am talking about : private void button1_Click( object sender, EventArgs e) { for ( int i = 0; i { System.Random rand = new System.Random(); int number = rand.next(180); comboBox1.Items.Add(number.ToString( "D2" )); } } //this was original; //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")); //} //}

Amit Gupta · Answer

1. Generate random numbers using random.Next(..) using Random class object and put it in a variable. 2. Change its sequence (as required) 3. Finally bind this numbers to combobox

Rajkiran Swain · Answer

https://answers.unity.com/questions/1505476/generate-a-random-number-in-a-range-based-on-dropd.html