Laravel 6 API RESTful，过滤了多对多关系

Question

我正在使用 laravel 6 开发多语言 API，我的数据库中有这种情况：

Categories
id
other not relevants fields

Languages:
id
name
code

Category_Language
id
language_id
category_id
name --> this is the name of the category in the specific language.

现在我有 2 个模型，第一个用于类别

class Category extends Model
{    
    public function languages()
    {
        return $this->belongsToMany('App\Models\v1\Language')->withTimestamps()->withPivot('name');
    }
}

和第二个模型

class Language extends Model
{
    protected $fillable = ['code', 'name', 'image_id', 'enabled'];

    public function categories() {
        return $this->belongsToMany('App\Models\v1\Category')->withTimestamps()->withPivot('name');
    }
}

在我的逻辑中（我正在使用服务模式），我在创建类别时使用了这种方法，每次传递这样的 JSON 对象时：

{
  "names": [
    {
      "languageId": 1,
      "name": "Hello"
    },
    {
      "languageId": 2,
      "name": "Hola"
    }
  ]
}

首先我创建一个类别（验证语言的 id 是否真的存储在数据库中），然后使用多对多的 laravel 功能，我将语言和名称附加到类别模型中，如下所示：

 foreach($request->names as $name) {
    $category->languages()->attach($name['languageId'], ['name' => $name['name']]);
 }

现在这似乎工作得很好，不用像这样使用 API 资源过滤语言就可以检索所有类别：

public function toArray($request)
{
    $category = [];
    $category['id'] = $this->id;
    $category['languages'] = [];

    $category['languages'] = $this->languages->map(function ($language) {
        return [
            'languageId' => $language->id,
            'languageCode' => $language->code,
            'languageName' => $language->name,
            'categoryName' => $language->pivot->name,
        ];
    });

    return $category;
}

这是具有 3 种语言的 3 个类别的输出（伪造数据）：

array:3 [
  "data" => array:3 [
    0 => array:4 [
      "id" => 1
      "languages" => array:3 [
        0 => array:4 [
          "languageId" => 1
          "languageCode" => "pa"
          "languageName" => "Kacey Trantow"
          "categoryName" => "quasi"
        ]
        1 => array:4 [
          "languageId" => 2
          "languageCode" => "ne"
          "languageName" => "Mr. Alexandre Heathcote"
          "categoryName" => "perferendis"
        ]
        2 => array:4 [
          "languageId" => 3
          "languageCode" => "kj"
          "languageName" => "Mr. Misael Robel"
          "categoryName" => "repudiandae"
        ]
      ]
      "imageUrl" => null
      "enabled" => true
    ]
    1 => array:4 [
      "id" => 2
      "languages" => array:3 [
        0 => array:4 [
          "languageId" => 1
          "languageCode" => "pa"
          "languageName" => "Kacey Trantow"
          "categoryName" => "non"
        ]
        1 => array:4 [
          "languageId" => 2
          "languageCode" => "ne"
          "languageName" => "Mr. Alexandre Heathcote"
          "categoryName" => "vitae"
        ]
        2 => array:4 [
          "languageId" => 3
          "languageCode" => "kj"
          "languageName" => "Mr. Misael Robel"
          "categoryName" => "suscipit"
        ]
      ]
      "imageUrl" => null
      "enabled" => true
    ]
    2 => array:4 [
      "id" => 3
      "languages" => array:3 [
        0 => array:4 [
          "languageId" => 1
          "languageCode" => "pa"
          "languageName" => "Kacey Trantow"
          "categoryName" => "molestiae"
        ]
        1 => array:4 [
          "languageId" => 2
          "languageCode" => "ne"
          "languageName" => "Mr. Alexandre Heathcote"
          "categoryName" => "esse"
        ]
        2 => array:4 [
          "languageId" => 3
          "languageCode" => "kj"
          "languageName" => "Mr. Misael Robel"
          "categoryName" => "beatae"
        ]
      ]
      "imageUrl" => null
      "enabled" => true
    ]
  ]

现在，问题是当我想过滤仅传递 1 种特定语言的类别时，这是一个经典用例，当用户在导航期间仅使用 1 种语言时，如果我只想要 1 种语言（以及 1 个名称在类别）我需要什么样的操作？

我已经用特定的过滤器组织了服务，并为任何集合排序，但这种需求似乎让我抓狂！

所以这是我构建过滤器、排序和动态包含的最终 getCategories 方法：

-

想像得到这样的查询字符串：

http://localhost:8000/api/v1/categories?orderBy=id:asc&include=language&language.code=EN

我想要这样的回应：

array:3 [
      "data" => array:3 [
        0 => array:4 [
          "id" => 1
          "languages" => array:3 [
            0 => array:4 [
              "languageId" => 1
              "languageCode" => "EN"
              "languageName" => "English"
              "categoryName" => "quasi"
            ]
          ]
        ]
        1 => array:4 [
          "id" => 2
          "languages" => array:3 [
            0 => array:4 [
              "languageId" => 1
              "languageCode" => "EN"
              "languageName" => "English"
              "categoryName" => "non"
            ]
          ]
        ]
        2 => array:4 [
          "id" => 3
          "languages" => array:3 [
            0 => array:4 [
              "languageId" => 1
              "languageCode" => "EN"
              "languageName" => "English"
              "categoryName" => "molestiae"
            ]
          ]
        ]
      ]

我需要这样的东西，但一般来说，因为我们有其他实体使用相同的多语言系统逻辑。

为了澄清我想要的是像这样的连接查询：

SELECT *
FROM categories ca INNER JOIN category_language cl ON ca.id =cl.category_id
                 INNER JOIN languages lan ON lan.id = cl.language_id
WHERE lan.code = 'EN'; 

您将仅检索英语语言的类别列表...

感谢您的任何建议和帮助！

Answer 1

我认为这并不是最优雅的方法，但我已经通过一些解决方法解决了。

首先，我不在查询字符串中传递语言代码，而是在使用 X 本地化参数的 Header 参数中传递，就像在这个 tutorial for laravel with multilanguages 中解释的那样。

其次，当检索所有资源时，我总是在服务方法中返回 query->paginate()：

public function getCategories(Request $request)
{
    $sort = $this->buildSort($request->sort ?? '', 'id', 'asc');
    $where = $this->buildWhere($request->where ?? '');
    $includes = $this->buildWith($request->include ?? '');
    $query = Category::orderBy($sort[0], $sort[1]);

    if (!empty($include)) {
        $query = $query->with($includes);
    }

    if (!empty($where)) {
        $query = $query->where($where);
    }

    return $query->paginate();

}

最后，我将请求标头截获到 Category API 资源中，如果设置了 X-localization 参数，我会在地图中过滤语言数组，如下所示：

public function toArray($request)
{
    $languageCode = $request->header('X-localization') ?? null;

    $category = [];
    $category['id'] = $this->id;
    $category['languages'] = [];

    $languages = $this->languages;

    if ($languageCode) {
        $languages = $languages->where('code', '=', $languageCode);
    }

    $languages = $languages->map(function ($language) use ($languageCode, $category) {
        return [
            'languageId' => $language->id,
            'languageCode' => $language->code,
            'languageName' => $language->name,
            'categoryName' => $language->pivot->name,
        ];
    });

    $category['languages'] = $languages;


    return $category;

}

结果正是我想要的，没有在我的项目中使用纯 SQL（假数据）：

array:3 [
  "data" => array:3 [
    0 => array:4 [
      "id" => 1
      "languages" => array:1 [
        0 => array:4 [
          "languageId" => 1
          "languageCode" => "ho"
          "languageName" => "Dott. Orfeo Sartori"
          "categoryName" => "rem"
        ]
      ]
      "imageUrl" => null
      "enabled" => true
    ]
    1 => array:4 [
      "id" => 2
      "languages" => array:1 [
        0 => array:4 [
          "languageId" => 1
          "languageCode" => "ho"
          "languageName" => "Dott. Orfeo Sartori"
          "categoryName" => "accusamus"
        ]
      ]
      "imageUrl" => null
      "enabled" => true
    ]
    2 => array:4 [
      "id" => 3
      "languages" => array:1 [
        0 => array:4 [
          "languageId" => 1
          "languageCode" => "ho"
          "languageName" => "Dott. Orfeo Sartori"
          "categoryName" => "totam"
        ]
      ]
      "imageUrl" => null
      "enabled" => true
    ]
  ]
  "links" => array:4 [
    "first" => "http://localhost/api/v1/categories?page=1"
    "last" => "http://localhost/api/v1/categories?page=1"
    "prev" => null
    "next" => null
  ]
  "meta" => array:7 [
    "current_page" => 1
    "from" => 1
    "last_page" => 1
    "path" => "http://localhost/api/v1/categories"
    "per_page" => 15
    "to" => 3
    "total" => 3
  ]
]

进程以退出代码 1 结束

Answer 2

你可以像这样加入雄辩：

$language_code=$request->input('language_code');

    $values=Category::query()->join('category_language','category_language.category_id','=','categories.id')
->join('languages','languages.id','=','category_language.language_id')
->where('languages.code','=',$language_code)
->select('categories.*','languages.code')
    ->orderBy('categories.id')->get();