2017-03-01 50 views
2

轉換這些我的表數據庫模型sub_bidang我想顯示返回數據,以JSON

id id_bidang  name  
1  1   Backend Developer 
2  1   Frontend Developer 
3  2   Senior Marketing 

我想通過字段ID bidang在陣列NAMA子bidang來顯示所有的名字

$data = Posting::find($id); 

    foreach ($data->sub_bidang as $value) { 

     $data = [ 

      'id posting job' => $data->id_posting_job, 
      'id kategori' => $data->id_kategori, 
      'nama kategori' => $data->kategori_posting['nama'], 
      'id bidang' => $data->id_bidang, 
      'nama bidang' => $data->bidang['nama'], 
      'nama sub bidang' => [ 
       'bidang' => $value->nama 
      ] 
     ]; 

     return response()->json($data); 
    } 

但無法正常工作,只顯示數組nama sub bidang中的後端開發人員。這樣

{ 
    "id posting job": 1, 
    "id kategori": 1, 
    "nama kategori": "Part Time", 
    "id bidang": 1, 
    "nama bidang": "IT", 
    "nama sub bidang": { 
    "bidang": "Backend Developer" 
    } 
} 

應該有顯示後端開發者和開發前端, 什麼問題?

+0

右語法這裏:'返回響應 - > JSON([ 「數據」=>「返回任何值或在此定義陣列]);' –

回答

2

在這裏嘗試以下

....................... 
....................... 
$result_data = array(); 
foreach ($data->sub_bidang as $value) { 

    $result_data[] = array(
     'id posting job' => $data->id_posting_job, 
     'id kategori' => $data->id_kategori, 
     'nama kategori' => $data->kategori_posting['nama'], 
     'id bidang' => $data->id_bidang, 
     'nama bidang' => $data->bidang['nama'], 
     'nama sub bidang' => array(
      'bidang' => $value->nama 
     ) 
    ); 


} 
return response()->json($result_data); 

注:$result_date是PHP數組所以response()->json()方法會自動生成JSON

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