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 $pass;
}
当前回答
如果你在PHP7上，你可以使用random_int()函数:
function generate_password($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 $str;
}
其他回答
如果你在PHP7上，你可以使用random_int()函数:
function generate_password($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;
}
下面是另一个密码生成器代码片段。 控制长度，数字和特殊字符计数和列表。
其他解决方案的一个问题是，它们没有包含重复字符的选项。而下面的脚本也可以做到这一点。
$length = random_int(30, 40);
$pass = [];
$lowers = range('a', 'z');
$uppers = range('A', 'Z');
$digits = range('0', '9');
$specials = ['.', '-', '_', '^', '#', '(', ')'];
$specialCount = random_int(1, 5);
$digitCount = random_int(1, 9);
for ($i = 0; $i < $length - $specialCount - $digitCount; $i++) {
$pass[] = random_int(1, PHP_INT_MAX) % 2 == 0 ? $uppers[array_rand($uppers)] : $lowers[array_rand($lowers)];
}
for ($i = 0; $i < $specialCount; $i++) {
$pass[] = $specials[array_rand($specials)];
}
for ($i = 0; $i < $digitCount; $i++) {
$pass[] = $digits[array_rand($digits)];
}
shuffle($pass)
$pass = implode('', $pass);
//define a function. It is only 3 lines!
function generateRandomPassword($length = 5){
$chars = "0123456789bcdfghjkmnpqrstvwxyzBCDFGHJKLMNPQRSTVWXYZ";
return substr(str_shuffle($chars),0,$length);
}
//usage
echo generateRandomPassword(5); //random password legth: 5
echo generateRandomPassword(6); //random password legth: 6
echo generateRandomPassword(7); //random password legth: 7
这是基于本页的另一个答案，https://stackoverflow.com/a/21498316/525649
这个答案只生成十六进制字符，0-9,a-f。对于一些看起来不像hex的东西，试试这个:
str_shuffle(
rtrim(
base64_encode(bin2hex(openssl_random_pseudo_bytes(5))),
'='
).
strtoupper(bin2hex(openssl_random_pseudo_bytes(7))).
bin2hex(openssl_random_pseudo_bytes(13))
)
Base64_encode返回更广泛的字母数字字符 Rtrim有时会在结尾删除=
例子:
32 efvfgdg891be5e7293e54z1d23110m3zu3fmjb30z9a740ej0jz4 b280R72b48eOm77a25YCj093DE5d9549Gc73Jg8TdD9Z0Nj4b98760 051年b33654c0eg201cfw0e6na4b9614ze8d2fn49e12y0zy557aucb8 y67Q86ffd83G0z00M0Z152f7O2ADcY313gD7a774fc5FF069zdb5b7
对于为用户创建界面来说，这不是很可配置的，但对于某些目的来说，这是可以的。增加字符数，以弥补特殊字符的不足。
Generates a strong password of length 8 containing at least one lower case letter, one uppercase letter, one digit, and one special character. You can change the length in the code too. function checkForCharacterCondition($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();
