Question

我想使用preg_math函数验证输入的密码是否符合我的复杂性和长度规则：

if (!preg_match('/(?=.*[A-Za-z])(?=.*[0-9])(?=.*[^a-zA-Z0-9])/', $password)) {
    $errors[]  = 'Password allows alpha-numerical sumbols and should contain number and letters between 6-24 symbols';
}

但是此功能无法正常工作。由于我的密码应至少包含一个数字（0-9）和一个字母（az）或符号（，），_， - 介于6到20个符号之间。< / p>

如何编写正则表达式？谢谢！

Answer 1

您正在比较整个字符串使用strpos：http://php.net/manual/en/function.strpos.php

$password = "test!";
//Enter characters you want to check for
$characters = "/(!?=.*[A-Za-z])(?=.*[0-9])(?=.*[^a-zA-Z0-9])/";
//Split into array to check each character
$chars = str_split($characters);

//Set wrong password by default and check for correct characters
$correctPassword = false;

//Check each character
foreach($chars as $char){
    //If characters have been found, set password as correct
    if (strpos($password, $char))
    {
        $correctPassword = true;
    }
}

//Also check for length, equal or greater then 6, smaller or equal to 20
if ( ($correctPassword) && ( strlen($password) >= 6 )  && ( strlen($password) <= 20 ) )
{
    echo "OK";
}
else
{
    echo "Password Unsafe";
}

希望这会有所帮助:)
如果这是您搜索的答案，请标记为答案。

Answer 2

您需要添加锚点并限制字符数，这是一种完成工作的方法：

preg_match('/^(?=.*[A-Za-z])(?=.*[0-9])(?=.*[^a-zA-Z0-9]).{6,20}$/', $password)

验证密码遵循复杂性和长度规则

2 个答案: