在 laravel 5.2 中使用 mysql 和 php 嵌套 json答案

【问题标题】：Nested json using mysql and php in laravel 5.2在 laravel 5.2 中使用 mysql 和 php 嵌套 json
【发布时间】：2016-08-09 00:40:00
【问题描述】：

我在 Laravel 5.2 中有一个选择代码，现在我想向嵌套的 json 显示所选业务 ID 的详细信息。类别和商家名称是重复的，所以我不想重复这两列。

我试过了

$menuproducts = DB::select("SELECT 
                             business.name as bizname, 
                             category.name as categoryname, 
                             products.name as productname
                        FROM menu
                             INNER JOIN products ON products.id = menu.product_id
                             INNER JOIN category ON category.id = products.category_id
                             INNER JOIN business ON business.id = menu.business_id
                        WHERE menu.business_id = $id"
                            );

        return $menuproducts;

结果如下：

[{"bizname":"Tivoli","categoryname":"Pizza","productname":"Margarita"},{"bizname":"Tivoli","categoryname":"Pizza","productname":"Molino"},{"bizname":"Tivoli","categoryname":"Pije","productname":"Coca Cola"},{"bizname":"Tivoli","categoryname":"Fast Food","productname":"Hamburger"},{"bizname":"Tivoli","categoryname":"Pizza","productname":"Pizza Sendwich"},{"bizname":"Tivoli","categoryname":"Pije","productname":"Pepsi"}]

想法是通过这种方式获取列表：

{  
   "business":[  
      {  
         "bizname":"Tivoli",
         "category":{  
            "name":"Pizza",
            "id":"500920000",
            "products":[  
               {  
                  "productname":"Margarita",
                  "price":"250"
               },
               {  
                  "productname":"Tuna Pizza",
                  "price":"350"
               }
            ],
            "category":{  
               "name":"Fast Food",
               "id":"500920000",
               "products":[  
                  {  
                     "productname":"Tost",
                     "price":"60"
                  },
                  {  
                     "productname":"Hamburger",
                     "price":"80"
                  }
               ]
            }
         }
      }
   ]
}

【问题讨论】：

标签： php mysql json laravel-5.2

【解决方案1】：

我已经尝试过这段代码，它工作得很好。但现在我想在 json 中有列名。

$menuproducts = DB::select("SELECT 
                                            business.name as bizname, 
                                            category.name as categoryname, 
                                            products.name as productname,
                                            menu.product_id as productid,
                                            menu.price as productprice
                                    FROM menu
                                        INNER JOIN products ON products.id = menu.product_id
                                        INNER JOIN category ON category.id = products.category_id
                                        INNER JOIN business ON business.id = menu.business_id
                                    WHERE menu.business_id = $id"
                                );

                    $rows = json_decode(json_encode($menuproducts), true);
                    $res=array();

                    foreach($rows as $row)
                    {
                      $res[$row['bizname']][$row['categoryname']][$row['productname']][] = $row['productprice'];

                    }

                    return Response::json($res, 200, array(), JSON_PRETTY_PRINT);

【讨论】：