获得随机字符串,格式("str"+xxxx)

public string getRandomString(string name)
{

Random rad = new Random();
        int value = rad.Next(1000, 10000);
        string res = value.ToString();
        string result = name + (DateTime.Now.ToString("yyyyMMdd")) + res;
        return result;
}

 

posted @ 2023-03-01 15:23  热心网友大熊  阅读(14)  评论(0)    收藏  举报