我试图在PHP中创建一个随机字符串,我得到绝对没有输出:

<?php
    function RandomString()
    {
        $characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
        $randstring = '';
        for ($i = 0; $i < 10; $i++) {
            $randstring = $characters[rand(0, strlen($characters))];
        }
        return $randstring;
    }

    RandomString();
    echo $randstring;

我做错了什么?


当前回答

以前的答案会生成不安全或难以输入的密码。

这是安全的,并且提供了用户更有可能实际使用的密码,而不是因为一些薄弱的东西而被丢弃。

// NOTE: On PHP 5.x you will need to install https://github.com/paragonie/random_compat

/**
 * Generate a password that can easily be typed by users.
 *
 * By default, this will sacrifice strength by skipping characters that can cause
 * confusion. Set $allowAmbiguous to allow these characters.
 */
static public function generatePassword($length=12, $mixedCase=true, $numericCount=2, $symbolCount=1, $allowAmbiguous=false, $allowRepeatingCharacters=false)
{
  // sanity check to prevent endless loop
  if ($numericCount + $symbolCount > $length) {
    throw new \Exception('generatePassword(): $numericCount + $symbolCount are too high');
  }

  // generate a basic password with just alphabetic characters
  $chars  = 'qwertyupasdfghjkzxcvbnm';
  if ($mixedCase) {
    $chars .= 'QWERTYUPASDFGHJKZXCVBNML';
  }
  if ($allowAmbiguous) {
    $chars .= 'iol';
    if ($mixedCase) {
      $chars .= 'IO';
    }
  }

  $password = '';
  foreach (range(1, $length) as $index) {
    $char = $chars[random_int(0, strlen($chars) - 1)];

    if (!$allowRepeatingCharacters) {
      while ($char == substr($password, -1)) {
        $char = $chars[random_int(0, strlen($chars) - 1)];
      }
    }

    $password .= $char;
  }


  // add numeric characters
  $takenSubstitutionIndexes = [];

  if ($numericCount > 0) {
    $chars = '23456789';
    if ($allowAmbiguous) {
      $chars .= '10';
    }

    foreach (range(1, $numericCount) as $_) {
      $index = random_int(0, strlen($password) - 1);
      while (in_array($index, $takenSubstitutionIndexes)) {
        $index = random_int(0, strlen($password) - 1);
      }

      $char = $chars[random_int(0, strlen($chars) - 1)];
      if (!$allowRepeatingCharacters) {
        while (substr($password, $index - 1, 1) == $char || substr($password, $index + 1, 1) == $char) {
          $char = $chars[random_int(0, strlen($chars) - 1)];
        }
      }

      $password[$index] = $char;
      $takenSubstitutionIndexes[] = $index;
    }
  }

  // add symbols
  $chars = '!@#$%&*=+?';
  if ($allowAmbiguous) {
    $chars .= '^~-_()[{]};:|\\/,.\'"`<>';
  }

  if ($symbolCount > 0) {
    foreach (range(1, $symbolCount) as $_) {
      $index = random_int(0, strlen($password) - 1);
      while (in_array($index, $takenSubstitutionIndexes)) {
        $index = random_int(0, strlen($password) - 1);
      }

      $char = $chars[random_int(0, strlen($chars) - 1)];
      if (!$allowRepeatingCharacters) {
        while (substr($password, $index - 1, 1) == $char || substr($password, $index + 1, 1) == $char) {
          $char = $chars[random_int(0, strlen($chars) - 1)];
        }
      }

      $password[$index] = $char;
      $takenSubstitutionIndexes[] = $index;
    }
  }

  return $password;
}

其他回答

这是我简单的一行解决方案,以生成一个使用友好的随机密码,排除字符看起来像“1”和“l”,“O”和“0”,等等…这里是5个字符,但你可以很容易地改变它:

$user_password = substr(str_shuffle('abcdefghjkmnpqrstuvwxyzABCDEFGHJKMNPQRSTUVWXYZ23456789'),0,5);

这个问题有很多答案,但没有一个是利用加密安全伪随机数生成器(CSPRNG)的。

简单、安全、正确的答案是使用RandomLib,不要白费力气。

对于那些坚持发明自己的解决方案的人,PHP 7.0.0将为此目的提供random_int();如果你还在使用PHP 5。x,我们为random_int()写了一个PHP 5的polyfill,这样你甚至可以在升级到PHP 7之前使用新的API。

在PHP中安全地生成随机整数并不是一项简单的任务。在生产环境中部署自己开发的算法之前,您应该始终与常驻StackExchange密码学专家进行检查。

有了安全的整数生成器,使用CSPRNG生成随机字符串就像在公园里散步一样简单。

创建安全的随机字符串

/**
 * Generate a random string, using a cryptographically secure 
 * pseudorandom number generator (random_int)
 *
 * This function uses type hints now (PHP 7+ only), but it was originally
 * written for PHP 5 as well.
 * 
 * For PHP 7, random_int is a PHP core function
 * For PHP 5.x, depends on https://github.com/paragonie/random_compat
 * 
 * @param int $length      How many characters do we want?
 * @param string $keyspace A string of all possible characters
 *                         to select from
 * @return string
 */
function random_str(
    int $length = 64,
    string $keyspace = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ'
): string {
    if ($length < 1) {
        throw new \RangeException("Length must be a positive integer");
    }
    $pieces = [];
    $max = mb_strlen($keyspace, '8bit') - 1;
    for ($i = 0; $i < $length; ++$i) {
        $pieces []= $keyspace[random_int(0, $max)];
    }
    return implode('', $pieces);
}

用法:

$a = random_str(32);
$b = random_str(8, 'abcdefghijklmnopqrstuvwxyz');
$c = random_str();

演示:https://3v4l.org/IMJGF(忽略PHP 5失败;它需要random_compat)

下面是一个简单的单行程序,它生成一个真正的随机字符串,而不需要任何脚本级循环或使用OpenSSL库。

echo substr(str_shuffle(str_repeat('0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ', mt_rand(1,10))), 1, 10);

把它分解,这样参数就清楚了

// Character List to Pick from
$chrList = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';

// Minimum/Maximum times to repeat character List to seed from
$chrRepeatMin = 1; // Minimum times to repeat the seed string
$chrRepeatMax = 10; // Maximum times to repeat the seed string

// Length of Random String returned
$chrRandomLength = 10;

// The ONE LINE random command with the above variables.
echo substr(str_shuffle(str_repeat($chrList, mt_rand($chrRepeatMin,$chrRepeatMax))), 1, $chrRandomLength);

此方法的工作原理是随机重复字符列表,然后打乱组合的字符串,并返回指定的字符数。

您可以进一步随机化它,通过随机化返回字符串的长度,将$chrRandomLength替换为mt_rand(8,15)(用于8到15个字符之间的随机字符串)。

实现这个函数的更好方法是:

function RandomString($length) {
    $keys = array_merge(range(0,9), range('a', 'z'));

    $key = "";
    for($i=0; $i < $length; $i++) {
        $key .= $keys[mt_rand(0, count($keys) - 1)];
    }
    return $key;
}

echo RandomString(20);

mt_rand在PHP 7中更加随机。rand函数是mt_rand的别名。

Laravel解决方案: 如果你正在使用Laravel框架,你可以像这样使用它的StringHelper。

使用“照亮\支持\斯特”;

Str::random($strlentgh = 16)