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
public static class StringHelper
{
private static readonly Random random = new Random();
private const int randomSymbolsDefaultCount = 8;
private const string availableChars = "ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789";
private static int randomSymbolsIndex = 0;
public static string GetRandomSymbols()
{
return GetRandomSymbols(randomSymbolsDefaultCount);
}
public static string GetRandomSymbols(int count)
{
var index = randomSymbolsIndex;
var result = new string(
Enumerable.Repeat(availableChars, count)
.Select(s => {
index += random.Next(s.Length);
if (index >= s.Length)
index -= s.Length;
return s[index];
})
.ToArray());
randomSymbolsIndex = index;
return result;
}
}
其他回答
DTB解决方案的一个稍微干净的版本。
var chars = "ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789";
var random = new Random();
var list = Enumerable.Repeat(0, 8).Select(x=>chars[random.Next(chars.Length)]);
return string.Join("", list);
您的风格偏好可能会有所不同。
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]
Eric J.写的代码很潦草(很明显这是6年前写的……他今天可能不会写那个代码),甚至还有一些问题。
与目前提出的一些替代方案不同,这个方案在密码学上是合理的。
不真实的…在密码中有一个偏差(正如在注释中所写的那样),bcdefgh比其他的更有可能(a不是,因为通过GetNonZeroBytes,它不会生成值为0的字节,因此a的偏差由它平衡),所以它在密码学上并不可靠。
这应该可以纠正所有的问题。
public static string GetUniqueKey(int size = 6, string chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890")
{
using (var crypto = new RNGCryptoServiceProvider())
{
var data = new byte[size];
// If chars.Length isn't a power of 2 then there is a bias if
// we simply use the modulus operator. The first characters of
// chars will be more probable than the last ones.
// buffer used if we encounter an unusable random byte. We will
// regenerate it in this buffer
byte[] smallBuffer = null;
// Maximum random number that can be used without introducing a
// bias
int maxRandom = byte.MaxValue - ((byte.MaxValue + 1) % chars.Length);
crypto.GetBytes(data);
var result = new char[size];
for (int i = 0; i < size; i++)
{
byte v = data[i];
while (v > maxRandom)
{
if (smallBuffer == null)
{
smallBuffer = new byte[1];
}
crypto.GetBytes(smallBuffer);
v = smallBuffer[0];
}
result[i] = chars[v % chars.Length];
}
return new string(result);
}
}
一种简单且高度安全的方法可能是生成加密Aes密钥。
public static string GenerateRandomString()
{
using Aes crypto = Aes.Create();
crypto.GenerateKey();
return Convert.ToBase64String(crypto.Key);
}
不使用Random的解决方案:
var chars = Enumerable.Repeat("ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789", 8);
var randomStr = new string(chars.SelectMany(str => str)
.OrderBy(c => Guid.NewGuid())
.Take(8).ToArray());
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