Laravel 关系：从其他表中添加一个 id

Question

public function show(Request $request, $slug = null)
{
    $products = Food::where("slug_food", $slug)->get();
    $relateds = Food::orderBy("created_at", "desc")->paginate(5);
    $ratings = Rating::where("food_id", $products[0]->id)->count();
    $comments = Rating::where("food_id", $products[0]->id)->get();

    if(count($products)>0){
        return view("user.product", compact("products", "relateds", "ratings", "comments"));
    }else{
        abort(404);
    }
    
}

public function addComment(Request $request, $slug = null)
{    
    $products = Food::where("id", $slug)->get();
    $ratings = new Rating();
    $ratings->comment = $request->input('comment');
    $ratings->food_id = $products;
    Auth::user()->ratings()->save($ratings);
       
}

错误代码： 无效的日期时间格式：1366 不正确的整数值：'[]' 列 efoodies.ratings.food_id 在第 1 行（SQL：插入到 ratings（comment，food_id，user_id， updated_at, created_at) 值 (dsa, [], 2, 2020-10-30 09:07:37, 2020-10-30 09:07:37))

我想插入一个餐桌食品的 id 并将其插入到评级表中

Answer 1

根据您的查询，我建议 FOOD 中的 ID 是您的 slug。

如果定义中是这种情况，您可以直接设置 Food 实体本身

public function addComment(Request $request, Food $food)
{    
    $ratings = new Rating();
    $ratings->comment = $request->input('comment');
    $ratings->food_id = $food->id;
    Auth::user()->ratings()->save($ratings);
       
}

如果不是..

public function addComment(Request $request, $slug = null)
{    
    $food = Food::where("slug", $slug)->firstOrFail();
    $ratings = new Rating();
    $ratings->comment = $request->input('comment');
    $ratings->food_id = $food->id;
    Auth::user()->ratings()->save($ratings);
       
}

PS.. 使用最后一行 Auth::user()->ratings()->save($ratings) 听起来不太好，而是我会设置

public function addComment(Request $request, $slug = null)
{    
    $food = Food::where("slug", $slug)->firstOrFail();
    $ratings = new Rating();
    $ratings->comment = $request->input('comment');
    $ratings->food_id = $food->id;
    $ratings->user_id = Auth::user()->id;
    $ratings->save();
}

在你的路线上小心使用？

Route::post('welcome/product/{slug?}', [LoginController::class, 'addComment'])->name("addComment"); 

这个蛞蝓？可能会导致空 slug 以及逻辑如何失败。 它必须没有？以确保只有有效的请求才会包含该 $slug。

Route::post('welcome/product/{slug}', [LoginController::class, 'addComment'])->name("addComment"); 

如果这个 $slug 等于不是“slug”的 Food 字段，只需在我的答案的第一行用正确的名称替换“slug”。