克隆不同查詢的雄辯模型

Question

很抱歉修改此問題（對所有已在此處回答的人）。 這個問題實際上是錯誤的。

$query = User::find(1)->books();  //books() is "return $this->hasMany('Book');"

$query_for_counting_history_books = clone $query;
$query_for_counting_geography_books = clone $query;

$number_of_history_books = $query_for_counting_history_books->where('type', '=', 'history')->count();
$number_of_geography_books = $query_for_counting_geography_books->where('type', '=', 'geography')->count();

$books = $query->paginate(10);

我希望查詢$ books：

SELECT * FROM books WHERE user_id = 1 limit 10;

但它的輸出是：

SELECT * FROM books WHERE user_id = 1 and type =“history”and type =“geography”limit 10;

• 我在這里錯過了一些關於克隆的事嗎？
• 該解決方案應該做些什么？

Answer 1

不確定為什么你的clone()不工作 - 但試試這個：

$query = User::where('confirmed', '=', '1');
$query_for_counting_male = $query->replicate();
$query_for_counting_female = $query->replicate();

http://laravel.com/api/4.2/Illuminate/Database/Eloquent/Model.html#method_replicate

Answer 2

由於$ query是HasMany的實例，因此克隆應該如下：

$ query_for_counting_history_books = clone $ query-> getQuery（）;

$ query_for_counting_geography_books = clone $ query-> getQuery（）;

http://laravel.com/api/4.2/Illuminate/Database/Eloquent/Relations/HasOneOrMany.html#method_getQuery