Laravel API APP多对多关系，如何以JSON返回特定信息? [英] Laravel API APP Many-Many Relationship, how to return specific information in JSON?

问题描述

一段时间以来，我一直在试图解决这个问题.基本上我有2个模型"Recipe"，成分"和一个控制器"RecipeController". 我正在使用Postman来测试我的API.当我转到使用RecipeController @ getRecipe的获取路线时，返回值如下图所示:

I been trying to figure this out for some time now. Basically i got 2 models ' Recipe ', ' Ingredient ' and one Controller ' RecipeController ' . I'm using Postman to test my API. When i go to my get route which uses RecipeController@getRecipe, the return value is as per the pic below:

返回获取路线"

如果我希望获取路线的返回值在下图的格式中，我该如何实现?通过这种方式，我的意思是我不想看到配方:created_at列，updated_at列以及配料:枢轴信息列，只想要名称和金额列信息.

If i want the return value of the get route to be in the FORMAT of the below pic, how do i achieve this? By this i mean i don't want to see for the recipes: the created_at column, updated_at column and for ingredients: the pivot information column, only want name and amount column information.

返回所需的值格式

食谱模型:

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class Recipe extends Model
{
    protected $fillable = ['name', 'description'];

    public function ingredients()
    {
      return $this->belongsToMany(Ingredient::class, 
      'ingredient_recipes')->select(array('name', 'amount'));
    }
}

成分模型:

<?php

namespace App;

use Illuminate\Database\Eloquent\Model;

class Ingredient extends Model
{
    protected $fillable = ['name', 'amount'];
}

RecipeController

RecipeController

<?php

namespace App\Http\Controllers;


use App\Ingredient;
use App\Recipe;
use Illuminate\Http\Request;

class RecipeController extends Controller {

public function postRecipe(Request $request)
{
    $recipe = new Recipe();
    $recipe->name = $request->input('name');
    $recipe->description = $request->input('description');
    $recipe->save();

    $array_ingredients = $request->input('ingredients');
    foreach ($array_ingredients as $array_ingredient) {
        $ingredient = new Ingredient();
        $ingredient->name = $array_ingredient['ingredient_name'];
        $ingredient->amount = $array_ingredient['ingredient_amount'];
        $ingredient->save();
        $recipe->ingredients()->attach($ingredient->id);
    }

    return response()->json(['recipe' => $recipe . $ingredient], 201);
}

public function getRecipe()
{
    $recipes = Recipe::all();
    foreach ($recipes as $recipe) {
        $recipe = $recipe->ingredients;
    }
    $response = [
        'recipes' => $recipes
    ];
    return response()->json($response, 200);
}

API路由:

Route::post('/recipe', 'RecipeController@postRecipe')->name('get_recipe');
Route::get('/recipe', 'RecipeController@getRecipe')->name('post_recipe');

谢谢你们！

推荐答案

我认为您最好的解决方案是使用Transformer.使用当前的实现，我建议您仅提取循环中所需的字段，即:

I think your best solution is using Transformer. Using your current implementation what I would recommend is fetching only the needed field in your loop, i.e:

foreach ($recipes as $recipe) {
    $recipe = $recipe->ingredients->only(['ingredient_name', 'ingredient_amount']);
}

尽管上述方法可能有效，但是由于将要进行大量的迭代/循环轮询数据库，因此当前的实现存在问题，我建议您急于加载该关系. 但是出于这个问题，您只需要 Transformer .

While the above might work, yet there is an issue with your current implementation because there will be tons of iteration/loop polling the database, I would recommend eager loading the relation instead. But for the sake of this question, you only need Transformer.

使用作曲器composer require league/fractal安装转换器，然后可以在app目录下创建一个名为Transformers的目录.

Install transformer using composer composer require league/fractal Then you can create a directory called Transformers under the app directory.

然后创建一个名为RecipesTransformer的类，并使用以下代码进行初始化:

Then create a class called RecipesTransformer, and initialize with:

namespace App\Transformers;

use App\Recipe;

use League\Fractal\TransformerAbstract;

class RecipesTransformer extends TransformerAbstract
{
    public function transform(Recipe $recipe)
    {
        return [
            'name' => $recipe->name,
            'description' => $recipe->description,
            'ingredients' => 
                $recipe->ingredients->get(['ingredient_name', 'ingredient_amount'])->toArray()
        ];
    }
}

然后，您可以像下面这样在控制器方法中使用此转换器:

Then you can use this transformer in your controller method like this:

use App\Transformers\RecipesTransformer;
......
public function getRecipe()
{
     return $this->collection(Recipe::all(), new RecipesTransformer);
     //or if you need to get one
     return $this->item(Recipe::first(), new RecipesTransformer);
}

您可以参考一个很好的教程喜欢以获取更多灵感，或直接访问分形"页面以获取详细信息.

You can refer to a good tutorial like this for more inspiration, or simply go to Fractal's page for details.

更新

为了使Fractal集合正常工作，因为如果您的项目中包含Dingo API，那么我给出的示例将可以正常工作，您可以通过以下方式手动创建它:

In order to get Fractal collection working since the example I gave would work if you have Dingo API in your project, you can manually create it this way:

public function getRecipe()
{
    $fractal = app()->make('League\Fractal\Manager');
    $resource = new \League\Fractal\Resource\Collection(Recipe::all(), new RecipesTransformer);

     return response()->json(
        $fractal->createData($resource)->toArray());
}

如果您要制作一个Item而不是Collection，则可以使用new \League\Fractal\Resource\Item.我建议您安装Dingo API或遵循此简单的教程，以便在没有不必要的重复的情况下进行更整洁的处理

In case you want to make an Item instead of collection, then you can have new \League\Fractal\Resource\Item instead. I would recommend you either have Dingo API installed or you can follow this simple tutorial in order to have in more handled neatly without unnecessary repeatition

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