我如何在c#中生成一个随机的8个字符的字母数字字符串?
当前回答
现在是单行风格。
private string RandomName()
{
return new string(
Enumerable.Repeat("ABCDEFGHIJKLMNOPQRSTUVWXYZ", 13)
.Select(s =>
{
var cryptoResult = new byte[4];
using (var cryptoProvider = new RNGCryptoServiceProvider())
cryptoProvider.GetBytes(cryptoResult);
return s[new Random(BitConverter.ToInt32(cryptoResult, 0)).Next(s.Length)];
})
.ToArray());
}
其他回答
不是100%确定,因为我没有测试这里的每个选项,但在我测试的选项中,这个是最快的。用秒表计时,它显示9-10滴答,所以如果速度比安全更重要,试试这个:
private static Random random = new Random();
public static string Random(int length)
{
var stringChars = new char[length];
for (int i = 0; i < length; i++)
{
stringChars[i] = (char)random.Next(0x30, 0x7a);
return new string(stringChars);
}
}
您只需使用程序集SRVTextToImage。并编写下面的代码生成随机字符串。
CaptchaRandomImage c1 = new CaptchaRandomImage();
string text = c1.GetRandomString(8);
它主要用于实现验证码。但对你来说也一样。希望能有所帮助。
在这个线程中只是一些不同答案的性能比较:
方法与设置
// what's available
public static string possibleChars = "abcdefghijklmnopqrstuvwxyz";
// optimized (?) what's available
public static char[] possibleCharsArray = possibleChars.ToCharArray();
// optimized (precalculated) count
public static int possibleCharsAvailable = possibleChars.Length;
// shared randomization thingy
public static Random random = new Random();
// http://stackoverflow.com/a/1344242/1037948
public string LinqIsTheNewBlack(int num) {
return new string(
Enumerable.Repeat(possibleCharsArray, num)
.Select(s => s[random.Next(s.Length)])
.ToArray());
}
// http://stackoverflow.com/a/1344258/1037948
public string ForLoop(int num) {
var result = new char[num];
while(num-- > 0) {
result[num] = possibleCharsArray[random.Next(possibleCharsAvailable)];
}
return new string(result);
}
public string ForLoopNonOptimized(int num) {
var result = new char[num];
while(num-- > 0) {
result[num] = possibleChars[random.Next(possibleChars.Length)];
}
return new string(result);
}
public string Repeat(int num) {
return new string(new char[num].Select(o => possibleCharsArray[random.Next(possibleCharsAvailable)]).ToArray());
}
// http://stackoverflow.com/a/1518495/1037948
public string GenerateRandomString(int num) {
var rBytes = new byte[num];
random.NextBytes(rBytes);
var rName = new char[num];
while(num-- > 0)
rName[num] = possibleCharsArray[rBytes[num] % possibleCharsAvailable];
return new string(rName);
}
//SecureFastRandom - or SolidSwiftRandom
static string GenerateRandomString(int Length) //Configurable output string length
{
byte[] rBytes = new byte[Length];
char[] rName = new char[Length];
SolidSwiftRandom.GetNextBytesWithMax(rBytes, biasZone);
for (var i = 0; i < Length; i++)
{
rName[i] = charSet[rBytes[i] % charSet.Length];
}
return new string(rName);
}
结果
在LinqPad中测试。对于长度为10的字符串,生成:
from Linq = chdgmevhcy [10] from Loop = gtnoaryhxr [10] from Select = rsndbztyby [10] from GenerateRandomString = owyefjjakj [10] from securefastrrandom = VzougLYHYP [10] from securefastrrandom - nocache = oVQXNGmO1S [10]
性能数据会有细微的变化,偶尔NonOptimized会更快,有时ForLoop和GenerateRandomString会切换谁领先。
LinqIsTheNewBlack (10000x) = 96762 ticks elapsed (9.6762 ms) ForLoop (10000x) = 28970滴答流逝(2.897毫秒) ForLoopNonOptimized (10000x) = 33336滴答流逝(3.3336毫秒) 重复(10000x) = 78547滴答流逝(7.8547毫秒) GenerateRandomString (10000x) = 27416 tick elapsed (2.7416 ms) securefastrrandom (10000x) = 13176滴答流逝(5ms)最低[不同的机器] securefastrrandom - nocache (10000x) = 39541 ticks elapsed (17ms) low[不同的机器]
这是我从Dot Net Perls的Sam Allen那里偷来的一个例子
如果你只需要8个字符,那么在系统中使用Path.GetRandomFileName()。IO命名空间。Sam说使用“Path.”这里的GetRandomFileName方法有时更优越,因为它使用RNGCryptoServiceProvider来获得更好的随机性。然而,它被限制为11个随机字符。”
GetRandomFileName总是返回一个12个字符的字符串,第9个字符是句点。所以你需要去掉句点(因为这不是随机的),然后从字符串中取出8个字符。实际上,你可以只取前8个字符而不用考虑句点。
public string Get8CharacterRandomString()
{
string path = Path.GetRandomFileName();
path = path.Replace(".", ""); // Remove period.
return path.Substring(0, 8); // Return 8 character string
}
PS:谢谢,Sam
I was looking for a more specific answer, where I want to control the format of the random string and came across this post. For example: license plates (of cars) have a specific format (per country) and I wanted to created random license plates. I decided to write my own extension method of Random for this. (this is in order to reuse the same Random object, as you could have doubles in multi-threading scenarios). I created a gist (https://gist.github.com/SamVanhoutte/808845ca78b9c041e928), but will also copy the extension class here:
void Main()
{
Random rnd = new Random();
rnd.GetString("1-###-000").Dump();
}
public static class RandomExtensions
{
public static string GetString(this Random random, string format)
{
// Based on http://stackoverflow.com/questions/1344221/how-can-i-generate-random-alphanumeric-strings-in-c
// Added logic to specify the format of the random string (# will be random string, 0 will be random numeric, other characters remain)
StringBuilder result = new StringBuilder();
for(int formatIndex = 0; formatIndex < format.Length ; formatIndex++)
{
switch(format.ToUpper()[formatIndex])
{
case '0': result.Append(getRandomNumeric(random)); break;
case '#': result.Append(getRandomCharacter(random)); break;
default : result.Append(format[formatIndex]); break;
}
}
return result.ToString();
}
private static char getRandomCharacter(Random random)
{
string chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
return chars[random.Next(chars.Length)];
}
private static char getRandomNumeric(Random random)
{
string nums = "0123456789";
return nums[random.Next(nums.Length)];
}
}