使用白名单对用户输入进行清理

我有这段代码可以清理名为“用户名”的变量上的用户输入：

$username_clean = preg_replace( "/[^a-zA-Z0-9_]/", "", $_POST['username'] );

if (!strlen($username_clean)){

die("username is blank!");

我想对此页面上的每个输入执行相同的过程，但我有大约 12 个不同的输入，因为它是一个注册表单。有没有一种更简单的方法来清理和检查每个输入，而不是对每个输入应用 preg_replace() 和 if 语句？

如果您想清理其中的所有元素$_POST，那么您可以创建一个清理函数并将其应用于所有元素array_map:

$post_clean = array_map("sanitization_function", $_POST);

然后你可以通过访问你的变量$post_clean代替$_POST.

它看起来像：

function sanitize($dirty){ 
    return preg_replace( "/[^a-zA-Z0-9_]/", "", $dirty ); 
}

$cPOST = array_map("sanitize", $_POST);

if (!strlen($cPOST['username'])){ 
    die("username is blank!"); 
}

如果您只想清理一部分$_POST元素，你可以这样做：

$cPOST = array();
$sanitize_keys = array('username','someotherkeytosanitize');
foreach($_POST as $k=>$v)
{
    if(in_array($k, $sanitize_keys))
    {
        $cPOST[$k] = preg_replace( "/[^a-zA-Z0-9_]/", "", $v);
    }
    else
    {
        $cPOST[$k] = $v;
    }
}

尝试这个：

$cPOST = array();
$sanitize_keys = array('username','someotherkeytosanitize');
for($_POST as $k=>$v)
{
    if(in_array($k, $sanitize_keys))
    {
        $cPOST[$k] = preg_replace( "/[^a-zA-Z0-9_]/", "", $v);
        if(strlen($cPOST[$k]) == 0){ 
            die("%s is blank", $k);
        }
    }
    else
    {
        $cPOST[$k] = $v;
    }
}
# At this point, the variables in $cPOST are the same as $_POST, unless you 
# specified they be sanitized (by including them in the $sanitize_keys array.
# Also, if you get here, you know that the entries $cPOST that correspond
# to the keys in $sanitize_keys were not blank after sanitization.

只需确保将 $sanitize_keys 更改为您想要清理的任何变量（或 $_POST 键）的数组。