PHP 函数没有到达最内层的嵌套 if 块

Question

所以我做了一些条件测试，由于某种原因我无法让这个函数到达最里面的一组 if/else。

我希望有人能告诉我我做了什么？

function verifyStridersIdentity($post_name, $modpass) {
    //is the post name strider?
    //this part works
    if (strtolower($post_name) == 'strider') {
        //is the mod pass set?
        //this part ALSO works
        if (!empty($modpass)) {
            //modpass has some kind of value


            $staff_name = md5_decrypt($modpass, KU_RANDOMSEED);//is it actually Strider though?
            //this seems to be the block I am having trouble getting into?
            //I can not get a value of either of these next two return statements.
            if ($staff_name == $post_name) {
                return 'This Works, It\'s me!'; //this is Strider
            }
            else {
                return 'This is '.$staff_name.' attempting to be Strider. This has been logged.';
            }
        }
        else {
            return 'Anonymous Test';
        }
    }   
    else {
        return $post_name;
    }

}

我希望内联评论能很好地解释发生了什么，但如果没有，请向我询问更多信息。

Answer 1

当我从原始 class 调用函数时，我使用了以下代码：

$post_name = verifyStridersIdentity($post_name,$modpass);

问题是 $modpass 实际上并不存在，所以我发送的是空白。

因此，纠正电话应该是这样的：

    $modpass = '';
    if (isset($_POST['modpassword'])) {
        $modpass = $_POST['modpassword'];
    }
    $post_name = verifyStridersIdentity($post_name,$modpass);

PHP 函数没有到达最内层的嵌套 if 块

PHP function not reaching the innermost nested if block

php

conditional

if-statement

nested