按用户级别限制内容

Question

基本上，我试图限制登录用户所能看到的内容，这取决于他们的排名。内容有几行，都有不同的等级要求。如果用户没有所需的级别，他将无法查看该行。然而，在这里的问题是，如果一行比它下面的任何行都有更高的等级要求，并且用户没有该级别，那么下面的所有行也都是不可见的。

public function Categories() {
    global $Template, $CatArray;

    $CatArray = array();
    $PermissionTable = array();

    $QueryCat = $this->mysqli->query("SELECT * FROM categories ORDER BY id ASC");

    while($FetchCat = $QueryCat->fetch_array()) {
        $PermissionTable["category"] = array("id" => $FetchCat["id"]); // store category ID as an id sub-array
        $data = explode(",", $FetchCat["ranks"]); // the ranks row in the database contains all ranks that can see this category, so here we split them up.
        foreach($data as $number) {
            $PermissionTable["category"] += array(
                "rank" => $data // apply rank requirements in a sub-array again
            );
        }

        if(in_array($Template["user"]["user_group"], $PermissionTable["category"]["rank"])) { // here, if the users rank is in the rank sub-array, they will be able to see it
            $CatArray[] = $FetchCat;
        } else { // otherwise display nothing
            $CatArray[] = null;
        }
    }

    $Template["CatArray"] = $CatArray;
    return $CatArray;
}

更新:这就是我的意思

​

Answer 1

我做了一些重构，但本质上，您将使用不同的函数来查看用户是否可以看到某个类别：

public function Categories() {
    global $Template, $CatArray;

    $CatArray = array();
    $PermissionTable = array();

    $QueryCat = $this->mysqli->query("SELECT * FROM categories ORDER BY id ASC");

    while($FetchCat = $QueryCat->fetch_array()) {
        $categoryRanks = explode(",", $FetchCat["ranks"]); // the ranks row in the database contains all ranks that can see this category, so here we split them up.

        $userCategoriesPermitted = in_array($Template["user"]["user_group"], $categoryRanks); //here we check if user rank is inside category rank

        if($userCategoriesPermitted) {
            $CatArray[] = $FetchCat; //add to return array
        }
    }

    $Template["CatArray"] = $CatArray;
    return $CatArray;
}

但这只是反映了一个不遵循第一范式的糟糕的数据库设计