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 $pass;
}
Generates a strong password of length 8 containing at least one lower case letter, one uppercase letter, one digit, and one special character. You can change the length in the code too.
function checkForCharacterCondition($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();
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 $f(12)。PHP_EOL;”
这是来自https://stackoverflow.com/a/41077923/5599052的想法
这是基于本页的另一个答案,https://stackoverflow.com/a/21498316/525649
这个答案只生成十六进制字符,0-9,a-f。对于一些看起来不像hex的东西,试试这个:
str_shuffle(
rtrim(
base64_encode(bin2hex(openssl_random_pseudo_bytes(5))),
'='
).
strtoupper(bin2hex(openssl_random_pseudo_bytes(7))).
bin2hex(openssl_random_pseudo_bytes(13))
)
Base64_encode返回更广泛的字母数字字符
Rtrim有时会在结尾删除=
例子:
32 efvfgdg891be5e7293e54z1d23110m3zu3fmjb30z9a740ej0jz4
b280R72b48eOm77a25YCj093DE5d9549Gc73Jg8TdD9Z0Nj4b98760
051年b33654c0eg201cfw0e6na4b9614ze8d2fn49e12y0zy557aucb8
y67Q86ffd83G0z00M0Z152f7O2ADcY313gD7a774fc5FF069zdb5b7
对于为用户创建界面来说,这不是很可配置的,但对于某些目的来说,这是可以的。增加字符数,以弥补特殊字符的不足。
如果你在PHP7上,你可以使用random_int()函数:
function generate_password($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;
}