使用php從JSON對象中選擇數據

Question

我有json數據我想選擇一些數據，然後我告訴你請仔細閱讀數據。

{ 
    "response": { 
     "status": 1, 
     "httpStatus": 200, 
     "data": [{ 
      "offer_id": "6912", 
      "Thumbnail": { 
       "10116": { 
        "id": "10116", 
        "offer_id": "6912", 
        "display": "Icandytv_IN_Call-30-19-20-51).gif", 
        "thumbnail": "https:\/\/media.go2speed.org\/brand\/files\/mobvista\/6912\/thumbnails_100\/Icandytv_IN_Call(04-30-19-20-51).gif" 
       } 
      } 
     }], 
     "errors":[] , 
     "errorMessage": null 
    } 
} 

從以上數據我想收集縮略圖的價值普萊舍幫助我找到它使用PHP

Answer 1

也許這是一個複製/粘貼的問題，而是你的JSON字符串缺少一些}。但這不是問題。如果您要訪問thumbnail（小寫）在你的JSON，嘗試以下操作：

<?php 

$json = '{"response": {"status":1,"httpStatus":200,"data":[{"offer_id":"6912","Thumbnail":{"10116":{"id":"10116","offer_id":"6912","display":"Icandytv_IN_Call (04-30-19-20-51).gif","thumbnail":"https://media.go2speed.org/brand/files/mobvista/6912/thumbnails_100/Icandytv_IN_Call(04-30-19-20-51).gif"}}}],"errors":[],"errorMessage":null}}'; 

// Make JSON accessible for PHP 
$data = json_decode($json); 

// Get access to Thumbnail object 
$thumbnail0 = $data->response->data[0]->Thumbnail; 

// Get access to thumbnail object "10116" 
$thumbnail_10116 = $thumbnail0->{"10116"}; 

// Thumbnail Url 
$thumbnailUrl = $thumbnail_10116->thumbnail; 

echo $thumbnailUrl . "\n"; 

// or in one swoop 
$thumbnailUrl2 = $data->response->data[0]->Thumbnail->{"10116"}->thumbnail; 

echo $thumbnailUrl2 . "\n"; 

?> 

希望，幫助