如何從URL中獲取JSON參數

Question

你好，我來自Get JSON object from URL

我試圖得到marketCap-> usd。

誰能告訴我我做錯了嗎？爲echo $obj->marketCap->usd;

$ch = curl_init(); 
curl_setopt($ch, CURLOPT_SSL_VERIFYPEER, false); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true); 
curl_setopt($ch, CURLOPT_URL, 'http://coinmarketcap.northpole.ro/api/v5/ZCL.json'); 
$result = curl_exec($ch); 
curl_close($ch); 

$obj = json_decode($result); 
echo $result->marketCap->usd; 

Answer 1

變化echo $result->marketCap->usd;您儲存了$obj解碼的數據。

Answer 2

最後你的代碼行應該

echo $obj->marketCap->usd;