如何随机生成一个由16个数字组成的字符串

Question

GUID  ;() 你截取一部分 转成 ASCII

Answer 1

string strNewStr = "";
Random rd = new Random((int)System.DateTime.Now.Ticks);
for(int i = 0; i < 10; i++)
{
    strNewStr += (char)((int)'0' + rd.Next(9));
}

Answer 2

public string GetByRndNum()
    {
        string Vchar = "0,1,2,3,4,5,6,7,8,9";
        string[] VcArray = Vchar.Split(',');
        string VNum = "";        
        //采用一个简单的算法以保证生成随机数的不同
        Random rand = new Random();
        int[] rndArr = new int[rndLength];
        for (int i = 1; i < rndLength + 1; i++)
        {   
            if(i>1)
            {
                do
                {
                    rndArr[i-1] = rand.Next(16);//rand.Next(35) ;
                    bool goon = false;
                    for(int ix=0; ix<i-1; ix++)
                    {
                        if (rndArr[ix] == rndArr[i-1])
                        {
                            goon = true;
                            break;
                        }
                    }
                    if (!goon)
                        break;
                } while (1 == 1);
            }
            else
                rndArr[i-1] = rand.Next(16);
            VNum += rndArr[i-1];
        }
        return VNum;
    }

Answer 3

MessageBox .Show (DateTime .Now .ToString ("yyMMddhhmmssffff"));重复的机率很小了

Answer 4

随机两个2位数
a b
a*b+a-b
再算md5
重复几率小点

Answer 5

Random r = new Random();
string str = r.Next(10000000,99999999).ToString() + r.Next(10000000,99999999).ToString();

Answer 6

需要把Random声明为类成员，使用它连续生成随机数，不会重复~