这将创建一个20个字符的十六进制字符串:

$string = bin2hex(openssl_random_pseudo_bytes(10)); // 20 chars

在PHP 7 (random_bytes())中:

$string = base64_encode(random_bytes(10)); // ~14 characters, includes /=+
// or
$string = substr(str_replace(['+', '/', '='], '', base64_encode(random_bytes(32))), 0, 32); // 32 characters, without /=+
// or
$string = bin2hex(random_bytes(10)); // 20 characters, only 0-9a-f

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;
}

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)

function getRandomString($length) {
  $salt = array_merge(range('a', 'z'), range(0, 9));
  $maxIndex = count($salt) - 1;

  $result = '';
  for ($i = 0; $i < $length; $i++) {
    $index = mt_rand(0, $maxIndex);
    $result .= $salt[$index];
  }
  return $result
}

该函数的编辑版本工作正常，但我发现只有一个问题:您使用了错误的字符来包含$字符，因此'字符有时是生成的随机字符串的一部分。

要解决这个问题，请更改:

$characters = ’0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ’;

to:

$characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';

这种方法只使用包含的字符，'字符永远不会是生成的随机字符串的一部分。