Question

在Laravel中我有一个场景，其中不同的用户可以访问视图刀片，在那里他们可以看到他们创建的帖子。

我刚刚传递了所有数据，但我想知道如何根据用户将数据传递给视图。

例如，如果我是root用户，我会看到一切如此

Post::get()

然后

return view('someview', compact('post')

哪会返回帖子

基本上我尝试的是这样的......

if(user->role = their role) then you get query 1 else you get query 2

您认为使用条件查询范围是否可行？

更新

这是一个可怕的解决方案吗？

if($user->department == "Loans")
{
    echo "you are from loans FAM";
    $articles = Article::where('department', '=', 'Loans')->get();
} 
else if($user->department == "Digital")
{
    echo "you are from digital FAM";
    $articles = Article::where('department', '=', 'Digital')->get();
} 
else if($user->department == "Consulting")
{
    echo "you are from Consulting FAM";
    $articles = Article::where('department', '=', 'Consulting')->get();
}

Answer 1

如果您愿意，可以使用查询范围实现此目的。像这样：

class Post extends Model
{
    // ...

    public function scopeByUser($query, User $user)
    {
        // If the user is not an admin, show only posts they've created
        if (!$user->hasRole('admin')) {
            return $query->where('created_by', $user->id);
        }

        return $query;
    }
}

然后你可以像这样使用它：

$posts = Post::byUser($user)->get();

响应您的更新：

class Article extends Model
{
    // ...

    public function scopeByUser($query, User $user)
    {
        // If the user is not an admin, show articles by their department.
        // Chaining another where(column, condition) results in an AND in
        // the WHERE clause
        if (!$user->hasRole('admin')) {
            // WHERE department = X AND another_column = another_value
            return $query->where('department', $user->department)
                ->where('another_column', 'another_value');
        }

        // If the user is an admin, don't add any extra where clauses, so everything is returned.
        return $query;
    }
}

您可以采用与上述相同的方式使用此功能。

Article::byUser($user)->get();

Laravel - 查询取决于用户

1 个答案: