隨機化的數字沒有在C＃中重複它

Question

我想隨機9號在每個蹤跡這樣的隨機數將是一次嘗試一些不同的......這裏是我的代碼

  Random num1random = new Random(); 
     label1.Text = num1random.Next(1, 9).ToString(); 
     label2.Text = num1random.Next(1, 9).ToString(); 
     label3.Text = num1random.Next(1, 9).ToString(); 
     label4.Text = num1random.Next(1, 9).ToString(); 
     label5.Text = num1random.Next(1, 9).ToString(); 
     label6.Text = num1random.Next(1, 9).ToString(); 
     label7.Text = num1random.Next(1, 9).ToString(); 
     label8.Text = num1random.Next(1, 9).ToString(); 
     label9.Text = num1random.Next(1, 9).ToString(); 

Answer 1

你真的不希望9個隨機數，你真的想以隨機順序從1到9的數字：

Random r = new Random(); 
var numbers = Enumerable.Range(1,9) // create a sequence of the integers 1 through 9 
    .OrderBy(x => r.Next()) // randomize the order 
    .ToArray(); // turn the sequence into an array. 

// assign the numbers to the labels 
label1.Text = numbers[0]; 
... 
label9.Text = numbers[8]; 

Answer 2

Random.Next不保證返回9個不同的值。你應該創建價值清單，然後洗牌。

Random num1random = new Random(); 
var numbers = Enumerable.Range(1, 9).OrderBy(item => num1random.Next()).ToList(); 
label1.Text = numbers[0].ToString(); 
//other labels