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

Ömer Faruk Çalışkan

6y1.1kVisual Basic .NET

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"));
}
}

7 Replies

Know the answer? Post it — somebody with the same question will find it here.

Rajkiran Swain6y ago
https://answers.unity.com/questions/1505476/generate-a-random-number-in-a-range-based-on-dropd.html
Ravi Patel6y ago
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 < 4; i++)
{
System.Random rand = new System.Random();
int number = rand.Next(180);
comboBox1.Items.Add(number.ToString("D2"));
}
}
Laxmidhar Sahoo6y ago
hi
public static void genaraterandom()
       {
           var _random = new Random();

           for (int i = 0; i < 4; i++)
           {

               var r = _random.Next(100, 1000);
           }


       }
Thanks and regards
Amit Gupta6y ago
In c#, method name always starts with upper case, so Next(..) should starts with upper case
Amit Gupta6y ago
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")); }
Ömer Faruk Çalışkan6y ago
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 < 4; 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 < 4; i++)
//{
//int number = rand.Next(100); // random number between 0 and 99
//comboBox1.Items.Add(number.ToString("D2"));
//}
//}
Amit Gupta6y ago
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