如何将数组推入代码初始化数组的对象的json键/值对中

Question

我要达到的目的

因此，我试图实现的是将一个数组键/值对推入一个本身就是数组的JSON对象中，因此只有一个对象，而键值对中的一个包含一个数组。

这是我要尝试的内容作为JSON输出，另一个数据数组内的另一个data2数组。

   [
    {
    "data": {
        "temp_service_id": "3",
        "name": "health checkup",
        "price": "10000",
        "service_id": "41",
        "data2": [
             {
              "fees": "2000",
              "service_name": "Anaesthesiologist"
             },
             {
              "fees": "300",
              "service_name": "Andrologist"
             },
           ]
        },  
     }
   ] 

我尝试过的这是我的控制者：

        $where['subtype_id'] = $this->post('org_id');
        $where['is_active'] = 'Y';
        $table_package_master = 'package_master';
        $fetch_package_name = $this->$model_name->fetch($table_package_master,$where);
        $array = [];
        $array1 = [];
        $array2 = [];
        if($fetch_package_name){
            foreach($fetch_package_name as $row){
                $where_r['t1.package_num'] = $row->package_num;
                $where_r['t1.is_active'] = 'Y';
                $where_r['services.is_active'] = 'Y';
                $where_r['t4.is_active'] = 'Y';
                $fetch_packages1 = $this->$model_name->fetch_packages1($where_r);
                $array['data'] = $fetch_packages1;
                $fetch_packages = $this->$model_name->fetch_packages($where_r);
                foreach($fetch_packages as $row1){
                    $where_re['services.service_id'] = $row1->service_id;
                    $where_re['services.is_active'] = 'Y';
                    $where_re['template_services.is_active'] = 'Y';                          

                    $fetch_package_ser = $this->$model_name->fetch_service_details($where_re);
                    array_push($array1,$fetch_package_ser);
                }
            }
            $array['data2'] = $array1;
            $array3 = [$array];
            $this->response($array3);
          }

这是我的模特：

             function fetch($table,$where_con){
             $this->db->select('*')->from($table)->where($where_con);
              return $this->db->get()->result();
                }

                function fetch_packages1($where){
             $this->db->select('t3.temp_service_id,t4.name,t4.price,services.service_id');
              $this->db->from('package_services as t1');
             $this->db->join('services','services.service_id = t1.service_id', 'LEFT');
             $this->db->join('template_services as t3','t3.temp_service_id = services.temp_service_id' , 'LEFT');
             $this->db->join('package_master as t4', 't4.package_num = t1.package_num','LEFT');
             $this->db->where($where);
             $this->db->group_by('t1.package_num');
             return $this->db->get()->row();
           }

          function fetch_service_details($where){
           $this->db->select('services.price as fees,template_services.service_name');
           $this->db->from('services');
           $this->db->join('template_services','template_services.temp_service_id = 
           services.temp_service_id','LEFT');
           $this->db->where($where);
           return $this->db->get()->row();
            }

[基本上，我想做的是将数据对象放在$ array变量中，该数据对象是来自一个表的数据，现在我所做的是将来自另一个查询的数据推入$ array1最后，我所做的只是将array_push到$ array3中，将来自$ array1和$ array2的所有数据组合在一起，但是我的输出就是这个：

         [
            {
             "data": {
                "temp_service_id": "3",
                "name": "health checkup",
                "price": "10000",
                "service_id": "41"
                     },
             "data2":  [
                     {
               "fees": "2000",
               "service_name": "Anaesthesiologist"
                     },
                     {
               "fees": "300",
               "service_name": "Andrologist"
                     }
               ]
            }
        ]

我在这里想念什么？我真的很困惑，无法理解如何推送data2数组。提前谢谢：）

Answer 1

用$array['data2'] = $array1;用Controller $array->data['data2'] = $array1;重写一行。

将数组data2放入data索引。

工作中Demo

Answer 2

经过反复考虑，我终于想出了一个解决方案

     $fetch_package_name = $this->$model_name->fetch($table_package_master,$where);
        $array = [];
        $array1 = [];
        if($fetch_package_name){
            for($i=0;$i<sizeof($fetch_package_name); $i++){
                $where_r['t1.package_num'] = $fetch_package_name[$i]->package_num;
                $where_r['t1.is_active'] = 'Y';
                $where_r['services.is_active'] = 'Y';
                $where_r['t4.is_active'] = 'Y';
                $fetch_packages1 = $this->$model_name->fetch_packages1($where_r);

                $array[$i]['name'] = $fetch_package_name[$i]->name;
                $array[$i]['price'] = $fetch_package_name[$i]->price;
                $where_re['t1.package_num'] = $fetch_package_name[$i]->package_num;
                $where_re['t1.is_active'] = 'Y';
                $where_re['services.is_active'] = 'Y';
                $fetch_packages = $this->$model_name->fetch_packages2($where_re);
                for($j=0; $j<sizeof($fetch_packages); $j++){
                    $array1[$j]['service_name'] = $fetch_packages[$j]->service_name;
                    $array1[$j]['price'] = $fetch_packages[$j]->price;
                }
                $array[$i]['data2'] = $array1;
            }
            // $this->response($array);
            $array2['data'] = $array;
            $array2['status'] = 10007; 
            // $data_response['data'] = $array;
            $this->response($array2);
    }       

IT首先给了我想要的响应。\

  { 
     "data":[ 
           { 
            "name":"Full body checkup",
             "price":"5000",
                "data2":[ 
                    { 
                    "service_name":"Allergist",
                    "price":"23"
                    },
                    { 
                    "service_name":"Andrologist",
                    "price":"300"
                    }
                 ]
            } 
      ],
  }

干杯！快乐的编码:-)