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 $pass;
}
当前回答
Create a file with this code in it. Call it like in the comments. <?php /** * @usage : * include_once($path . '/Password.php'); * $Password = new Password; * $pwd = $Password->createPassword(10); * return $pwd; * */ class Password { public function createPassword($length = 15) { $response = []; $response['pwd'] = $this->generate($length); $response['hashPwd'] = $this->hashPwd( $response['pwd'] ); return $response; } private function generate($length = 15) { $chars = "abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789!@#$%^&*(){}/?,><"; return substr(str_shuffle($chars),0,$length); } private function hashPwd($pwd) { return hash('sha256', $pwd); } } ?>
其他回答
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 $f(12)。PHP_EOL;”
这是来自https://stackoverflow.com/a/41077923/5599052的想法
你需要strlen($alphabet)，而不是常量字母的计数(相当于'alphabet')。
然而，rand并不是一个适合于此目的的随机函数。它的输出可以很容易地预测，因为它隐含地以当前时间作为种子。此外，兰特是不加密安全的;因此，从输出中确定其内部状态相对容易。
相反，从/dev/urandom读取以获得加密随机数据。
如果你在PHP7上，你可以使用random_int()函数:
function generate_password($length = 20){
$chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'.
'0123456789`-=~!@#$%^&*()_+,./<>?;:[]{}\|';
$str = '';
$max = strlen($chars) - 1;
for ($i=0; $i < $length; $i++)
$str .= $chars[random_int(0, $max)];
return $str;
}
旧答案如下:
function generate_password($length = 20){
$chars = 'ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz'.
'0123456789`-=~!@#$%^&*()_+,./<>?;:[]{}\|';
$str = '';
$max = strlen($chars) - 1;
for ($i=0; $i < $length; $i++)
$str .= $chars[mt_rand(0, $max)];
return $str;
}
安全警告:rand()不是一个加密安全的伪随机数生成器。在其他地方寻找在PHP中生成加密安全的伪随机字符串的方法。
试试这个(使用strlen而不是count，因为count在字符串上总是1):
function randomPassword() {
$alphabet = 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ1234567890';
$pass = array(); //remember to declare $pass as an array
$alphaLength = strlen($alphabet) - 1; //put the length -1 in cache
for ($i = 0; $i < 8; $i++) {
$n = rand(0, $alphaLength);
$pass[] = $alphabet[$n];
}
return implode($pass); //turn the array into a string
}
Demo
Generates a strong password of length 8 containing at least one lower case letter, one uppercase letter, one digit, and one special character. You can change the length in the code too. function checkForCharacterCondition($string) { return (bool) preg_match('/(?=.*([A-Z]))(?=.*([a-z]))(?=.*([0-9]))(?=.*([~`\!@#\$%\^&\*\(\)_\{\}\[\]]))/', $string); } $j = 1; function generate_pass() { global $j; $allowedCharacters = '0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ~`!@#$%^&*()_{}[]'; $pass = ''; $length = 8; $max = mb_strlen($allowedCharacters, '8bit') - 1; for ($i = 0; $i < $length; ++$i) { $pass .= $allowedCharacters[random_int(0, $max)]; } if (checkForCharacterCondition($pass)){ return '<br><strong>Selected password: </strong>'.$pass; }else{ echo 'Iteration '.$j.': <strong>'.$pass.'</strong> Rejected<br>'; $j++; return generate_pass(); } } echo generate_pass();
