我试图在php中生成一个随机密码。

但是我得到的都是'a'返回类型是数组类型,我希望它是字符串。对如何修改代码有什么想法吗?

谢谢。

function randomPassword() {
    $alphabet = "abcdefghijklmnopqrstuwxyzABCDEFGHIJKLMNOPQRSTUWXYZ0123456789";
    for ($i = 0; $i < 8; $i++) {
        $n = rand(0, count($alphabet)-1);
        $pass[$i] = $alphabet[$n];
    }
    return $pass;
}

当前回答

我创建了一个更全面、更安全的密码脚本。这将创建两个大写字母、两个小写字母、两个数字和两个特殊字符的组合。总共8个字符。

$char = [range('A','Z'),range('a','z'),range(0,9),['*','%','$','#','@','!','+','?','.']];
$pw = '';
for($a = 0; $a < count($char); $a++)
{
    $randomkeys = array_rand($char[$a], 2);
    $pw .= $char[$a][$randomkeys[0]].$char[$a][$randomkeys[1]];
}
$userPassword = str_shuffle($pw);

其他回答

Base_convert (uniqid('pass', true), 10,36);

我。e0m6ngefmj4

EDIT

正如我在评论中提到的,长度意味着暴力攻击比定时攻击更有效,所以不必担心“随机生成器有多安全”。安全性,特别是对于这个用例,需要补充可用性,所以上面的解决方案对于所需的问题已经足够好了。

然而,以防你在搜索安全的随机字符串生成器时偶然发现了这个答案(我假设有些人已经基于响应),对于生成令牌之类的东西,以下是此类代码的生成器的样子:

function base64urlEncode($data) {
    return rtrim(strtr(base64_encode($data), '+/', '-_'), '=');
}

function secureId($length = 32) {

    if (function_exists('openssl_random_pseudo_bytes')) {
        $bytes = openssl_random_pseudo_bytes($length);
        return rtrim(strtr(base64_encode($bytes), '+/', '0a'), '=');
    }
    else { // fallback to system bytes

        error_log("Missing support for openssl_random_pseudo_bytes");

        $pr_bits = '';

        $fp = @fopen('/dev/urandom', 'rb');
        if ($fp !== false) {
            $pr_bits .= @fread($fp, $length);
            @fclose($fp);
        }

        if (strlen($pr_bits) < $length) {
            error_log('unable to read /dev/urandom');
            throw new \Exception('unable to read /dev/urandom');
        }

        return base64urlEncode($pr_bits);
    }
}

我创建了一个更全面、更安全的密码脚本。这将创建两个大写字母、两个小写字母、两个数字和两个特殊字符的组合。总共8个字符。

$char = [range('A','Z'),range('a','z'),range(0,9),['*','%','$','#','@','!','+','?','.']];
$pw = '';
for($a = 0; $a < count($char); $a++)
{
    $randomkeys = array_rand($char[$a], 2);
    $pw .= $char[$a][$randomkeys[0]].$char[$a][$randomkeys[1]];
}
$userPassword = str_shuffle($pw);

一个简单的代码应该是这样的:

function generatePassword($len){
    $az = range("a","z");
    $AZ = range("A","Z");
    $num = range(0,9);
    $password = array_merge($az,$AZ,$num);
    return substr(str_shuffle(implode("",$password)),0, $len);
}
// testing 
$generate = range(8,32);
foreach($generate as $g){
    print "Len:{$g} = " . generatePassword($g)."\n";
}

输出:

Len:8 = G5uFhPKS
Len:9 = aU9x2NjvI
Len:10 = lJE9kxy3oD
Len:11 = tVh2CmpMdHW
Len:12 = ToXYHCPb58Ar
Len:13 = KIFVoLg5NdDzX
Len:14 = eFUabML28tXhf0
Len:15 = iegDCQcIMaxH0ST
Len:16 = sRvDmPo5IkaMqNO0
Len:17 = T5rwVDs6XGAqSU9KN
Len:18 = QwROWAfh1lpoCSaX0H
Len:19 = HP0trD4B9SQeUkNuAGV
Len:20 = P9Fdwqmu782ARHDiKGZM
Len:21 = 3Gxia9LPmCZM68dwe4YOf
Len:22 = ywFjuA2GDg0Oz8LVnCI94M
Len:23 = 16MiEVUgqPRueahlyvJfBz5
Len:24 = sPt0H9NSu5KrJTYeMXbOFgi7
Len:25 = QFKGTypaZlsMRnHPgNbVfIwxm
Len:26 = hbyJXtV81AEuMazS4GdFTINBUg
Len:27 = H3AiD95S4Z8xwMrz2L71GqUunaW
Len:28 = m8W2geIiO7Phc3H5Kyr1XCAs09Dv
Len:29 = MusNfYgOWnbrI62twRBvj38XEcDdi
Len:30 = VgNeILaRT2wvb4J7hzCMSHsquUBtnA
Len:31 = nhUvCxgOS94dsYjzBtcaTou1WIArMQP
Len:32 = AFSVQqCijuPMp0cGJNdDtzYX78erKB9w

快速。简单,干净和一致的格式,如果这是你想要的

$pw = chr(mt_rand(97,122)).mt_rand(0,9).chr(mt_rand(97,122)).mt_rand(10,99).chr(mt_rand(97,122)).mt_rand(100,999);

有一个简短的解决方案(php 8.1):

$dict = array_merge(
    ...array_map(
        fn(array $d): array => range(ord($d[0]), ord($d[1])),
        [["0", "9"], ["a", "z"], ["A", "Z"]]
    )
); 

$f = fn (int $len): string =>
    join(
        "",
        array_map(
            fn (): string => chr($dict[random_int(0, count($dict) - 1)]),
            range(0, $len)
        )
    ); 

echo $f(12) . PHP_EOL;

一行bash脚本:

PHP -r '$dict = array_merge(…到fn(数组$ d):数组= >范围(奥德($ d[0]),奥德($ d[1])),(“0”,“9”,“一个”、“z”,[“一”、“z”]]));$ f = fn (int len美元):字符串= >加入(“”,到(fn():字符串= >科($ dict [random_int (0, count ($ dict) - 1))),范围(0,len美元)));Echo $f(12)。PHP_EOL;”

这是来自https://stackoverflow.com/a/41077923/5599052的想法