2017-10-05 124 views
0

我想從這個字符串獲取圖像的URL http://www.example.com/xx.jpghttp://www.example.com/yy.jpg如何使用php獲取此字符串中的數據?

{ 
    "attachments": { 
     "data": [ 
     { 
      "subattachments": { 
       "data": [ 
        { 
        "description": "this is description", 
        "media": { 
         "image": { 
          "height": 720, 
          "src": "http://www.example.com/xx.jpg", 
          "width": 405 
         } 
        } 
        }, 
        { 
        "media": { 
         "image": { 
          "height": 720, 
          "src": "http://www.example.com/yy.jpg", 
          "width": 701 
         } 
        }, 
        "target": { 
         "id": "123456789", 
         "url": "http://www.example.com" 
        }, 
        "type": "text", 
        "url": "http://www.example.com" 
        } 
       ] 
      }, 
      "target": { 
       "id": "123456", 
       "url": "http://www.example.com" 
      }, 
      "title": "This is title", 
      "type": "number", 
      "url": "http://www.example.com" 
     } 
     ] 
    }, 
    "id": "123456789" 
} 

,所以我用這個代碼

$jason_data = file_get_contents('http://www.test.com'); 
$result_jason_data = json_decode($jason_data); 
echo $result_jason_data->attachments->data->media[0]->image; 
echo $result_jason_data->attachments->data->media[1]->image; 

但我沒有任何數據,你能告訴我,我該怎麼辦?

+5

的可能的複製[如何解析JSON(https://stackoverflow.com/questions/2591098/how-to-parse-json) –

+0

是的,這是可以做到一步一步,並檢查導致之間。我想你錯過了一個陣列。 –

+0

'echo $ result_jason_data-> attachments-> data [0] - > subattachments-> data [0] - > media-> image-> src;' –

回答

0
$array = json_decode($json); 
$xxImage = $array->attachments->data[0]->subattachments->data[0]->media->image->src; 
$yyImage = $array->attachments->data[0]->subattachments->data[1]->media->image->src; 
echo $xxImage; 
echo "<br>"; 
echo $yyImage; 
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