接收JSON响应（android）

我想发送一个带有信息电子邮件地址，名字和姓氏的JSON，并期望来自服务器的响应表示{status：“Created”}或{status：“Resend”}并根据答案会有一个弹出消息。我想知道我用什么从状态类中提取信息。谢谢！这里是我的代码接受接收JSON响应（android）

protected void sendJson(final String email, final String firstN, 
     final String lastN) { 
    Thread t = new Thread() { 
     public void run() { 
      Looper.prepare(); // For Preparing Message Pool for the child 
           // Thread 
      HttpClient client = new DefaultHttpClient(); 
      HttpConnectionParams.setConnectionTimeout(client.getParams(), 
        10000); // Timeout Limit 
      HttpResponse response; 
      JSONObject json = new JSONObject(); 
      try { 

       // post in the url 


       HttpPost post = new HttpPost(
         "https://iphone-radar.com/accounts"); 
       json.put("email_address", email); 
       json.put("first_name", firstN); 
       json.put("last_name", lastN); 
       StringEntity se = new StringEntity("JSON: " 
         + json.toString()); 
       se.setContentEncoding(new BasicHeader(HTTP.CONTENT_TYPE, 
         "application/json")); 
       post.setEntity(se); 
       response = client.execute(post); 
       /* Checking response */ 
       if (response != null) { 
        String str = response.getEntity().toString(); 
        if (str.equals("Created")) { 
         new AlertDialog.Builder(CreateAccount.this) 
           .setTitle("Account Creation Successful") 
           .setMessage(
             "An activation code has been sent to you. Please check your SPAM folder if you do not receive your activation code email") 
           .setNeutralButton("OK", null).show(); 
        } else if(str.equals("Resend")) { 
         new AlertDialog.Builder(CreateAccount.this) 
           .setTitle("Code Resent") 
           .setMessage(
             "Your activation code has been resent to your email.\n\nIf you are not receiving your activation code, our email is being blocked. Please email us at '[email protected]' and we will manually send you a code.") 
           .setNeutralButton("OK", null).show(); 
        } 


       } 
      } catch (Exception e) { 
       e.printStackTrace(); 
      } 

     }

来源

2011-08-04 Sean

你应该响应转换成字符串，然后创建一个JSONObject。然后你可以访问JSON对象的属性。试试这个：

org.json.JSONObject obj = new org.json.JSONObject(org.apache.http.util.EntityUtils.toString(response.getEntity())); 
        if ("Created".equals(obj.getString("status"))) { 
         new AlertDialog.Builder(CreateAccount.this) 
           .setTitle("Account Creation Successful") 
           .setMessage(
             "An activation code has been sent to you. Please check your SPAM folder if you do not receive your activation code email") 
           .setNeutralButton("OK", null).show(); 
        } else if("Resend".equals(obj.getString("status"))) { 
         new AlertDialog.Builder(CreateAccount.this) 
           .setTitle("Code Resent") 
           .setMessage(
             "Your activation code has been resent to your email.\n\nIf you are not receiving your activation code, our email is being blocked. Please email us at '[email protected]' and we will manually send you a code.") 
           .setNeutralButton("OK", null).show(); 
        }

来源

2011-08-04 00:46:48 Femi

谢谢你的提示，但是当我把它通过调试线路org.json.JSONObject的obj =新org.json.JSONObject（org.apache.http.util.EntityUtils.toString（ response.getEntity（）））;发送给我（异常e）我或服务器有问题吗？再次感谢 – Sean

它给了什么例外？使用'android.util.Log.v（“EXCEPTION”，“[”+ e.getMessage（）+“]”，e）;'在日志中打印出整个异常。要么是响应的实体为null（可能是服务器），要么在EntityUtils.toString方法中存在一些转换错误。 – Femi

它说[值为的java.lang.String类型不能转换为JSONObject]这是否意味着我使用了错误的网站？再次感谢 – Sean

接收JSON响应（android）

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