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

PHP生成随机字符的函数

这个PHP函数为我工作:

function cvf_ps_generate_random_code($length=10) {

   $string = '';
   // You can define your own characters here.
   $characters = "23456789ABCDEFHJKLMNPRTVWXYZabcdefghijklmnopqrstuvwxyz";

   for ($p = 0; $p < $length; $p++) {
       $string .= $characters[mt_rand(0, strlen($characters)-1)];
   }

   return $string;

}

用法:

echo cvf_ps_generate_random_code(5);

其他回答

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;
}

根据您的应用程序(我希望生成密码),您可以使用

$string = base64_encode(openssl_random_pseudo_bytes(30));

由于是base64,它们可以包含=或-以及所请求的字符。您可以生成一个更长的字符串,然后过滤和修剪它以删除它们。

Openssl_random_pseudo_bytes似乎是在php中生成适当随机数的推荐方法。为什么rand不用/dev/random,我不知道。

function randomString($length = 5) {
    return substr(str_shuffle(implode(array_merge(range('A','Z'), range('a','z'), range(0,9)))), 0, $length);
}

具体回答这个问题,有两个问题:

当你回显$randstring时,它不在作用域内。 字符在循环中没有连接在一起。

以下是更正后的代码片段:

function generateRandomString($length = 10) {
    $characters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
    $charactersLength = strlen($characters);
    $randomString = '';
    for ($i = 0; $i < $length; $i++) {
        $randomString .= $characters[random_int(0, $charactersLength - 1)];
    }
    return $randomString;
}

用下面的调用输出随机字符串:

// Echo the random string.
// Optionally, you can give it a desired string length.
echo generateRandomString();

请注意,这个答案的以前版本使用rand()而不是random_int(),因此生成可预测的随机字符串。因此,根据这个答案的建议,它被更改为更安全。

请尝试这个函数来生成一个自定义的随机字母数字字符串:

<?php
  function random_alphanumeric($length) {
    $chars = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ12345689';
    $my_string = '';
    for ($i = 0; $i < $length; $i++) {
      $pos = random_int(0, strlen($chars) -1);
      $my_string .= substr($chars, $pos, 1);
    }
    return $my_string;
  }
?>

你可以通过将字符串的长度传递给函数来调整结果,如下所示:

  $test_with_50_items = random_alphanumeric(50); // 50 characters
  echo $test_with_50_items;

示例(test_with_50_items): Y1FypdjVbFCFK6Gh9FDJpe6dciwJEfV6MQGpJqAfuijaYSZ86

如果你需要超过50个字符或更少,只需按你喜欢的方式调用函数:

  $test_with_27_items = random_alphanumeric(27); // 27 characters
  echo $test_with_27_items;

如果你需要两个或更多唯一的字符串,你可以使用while循环,这样你肯定会得到两个唯一的字符串…你可以用更多的弦做同样的事情,唯一的限制是你的幻想……

  $string_1 = random_alphanumeric(50);
  $string_2 = random_alphanumeric(50);
  while ($string_1 == $string_2) {
    $string_1 = random_alphanumeric(50);
    $string_2 = random_alphanumeric(50);
    if ($string_1 != $string_2) {
       break;
    }
  }
  echo $string_1;
  echo "<br>\n";
  echo $string_2;

$string_1: KkvUwia8rbDEV2aChWqm3AgeUZqyrRbUx2AxVhx5s4TSJ2VwA4

$string_2: XraO85YfxBBCInafvwipSOJwLmk6JMWiuWOxYQDnXohcn2D8K6

根据PHP 8.3, random_int()是“默认安全的”