2015-10-20 110 views
-2

我需要JSON格式响应这我得到的如何使用PHP SDK从twilio短信API

class Services_Twilio_Rest_Message 

我需要在以下JSON发送短信后返回的响应得到JSON格式响应:

{ 
    "sid": "sid", 
    "date_created": "Tue, 20 Oct 2015 06:01:14 +0000", 
    "date_updated": "Tue, 20 Oct 2015 06:01:14 +0000", 
    "date_sent": null, 
    "account_sid": "AccountSid", 
    "to": "+91999999999", 
    "from": "+18989898989", 
    "body": "I am trying to send a message having characters more than 160. But twilio allows concatenated messages upto 1600 characters length, i thought i should give it a try. So i wrote this weird message.", 
    "status": "queued", 
    "num_segments": "2", 
    "num_media": "0", 
    "direction": "outbound-api", 
    "api_version": "2010-04-01", 
    "price": null, 
    "price_unit": "USD", 
    "error_code": null, 
    "error_message": null, 
    "uri": "/2010-04-01/Accounts/AccountSid/Messages/SM745b97dcb56a4f9f82c52242ca3b5e92.json", 
    "subresource_uris": { 
    "media": "/2010-04-01/Accounts/AccountSid/Messages/SM745b97dcb56a4f9f82c52242ca3b5e92/Media.json" 
    } 
} 
+0

我不确定你在问什么。该API返回JSON,但PHP SDK将其封装在对象中,以便在应用程序中处理。 – philnash

回答

1

Ricky从Twilio在这里。

Services_Twilio_Rest_Message对象有一个toString method,它将返回json表示形式对象。如果你只是要输出的JSON用于调试你可以呼应回应你:

$message = $client->account->messages->create(array(
    "From" => "XXX-XXX-XXXX", 
    "To" => "XXX-XXX-XXXX", 
    "Body" => "Test message!", 
)); 

echo $message; 

如果你需要存储在一个变量,你可以在字符串显式转换$消息字符串:

$json = (string)$message; 

希望有帮助!

+0

这没有帮助 –

+0

对不起,在这种情况下,我可能不完全理解你的问题。 toString方法给你的json与你想要返回的东西之间有一个delta吗? – rickyrobinett