Loopj发布请求正文

Question

亲爱的我所有使用Loopj，真的很喜欢它。它让我的生活更轻松。现在我想要post json在发布请求的主体中。请检查它在做什么错我的代码在下面。

params.put("SaleOrderItems", mJsonArrayOfObject.toString()); 
    params.put("CustomerReferenceNumber", "asdf"); 
    // /*mSaleOrder.getmCustomerReferenceNo()*/); 
    params.put("RecordType", "HOS"); 
    params.put("DeliveryDate", "2012-12-28T12:04:27.3553985+01:00"); // mSaleOrder.getmDeliveryDate()); 
    params.put("SellToCustomerNumber", "user"); 

然后调用这个样子。

mAsyncHttpClient.post(WEBCONSTANTS.ORDER_SERVICE_SAVE_ORDER, mParams, 
      new AsyncHttpResponseHandler(){}; 

我得到这个错误

{"Message":"No HTTP resource was found that matches the request URI} 

请告诉我如何发送对象的JSON数组使用循环J POST请求的身体。 最好的问候，

Answer 1

我认为这是你在找什么：

String url = "<your site url>"; 
    JSONObject jdata = new JSONObject(); 
    try { 
     jdata.put("key1", val1); 
     jdata.put("key2", val2); 
    } catch (Exception ex) { 
     // json exception 
    } 
    StringEntity entity; 
    try { 
     entity = new StringEntity(jdata.toString()); 
     client.post(getBaseContext(), url, entity, "application/json", new AsyncHttpResponseHandler() { 
      @Override 
      public void onSuccess(String response) { 
       JSONObject res; 
       try { 
        res = new JSONObject(response); 
        Log.d("debug", res.getString("some_key")); // this is how you get a value out 
       } catch (JSONException e) { 
        // TODO Auto-generated catch block 
        e.printStackTrace(); 
       } 
      } 

     }); 
    } catch (UnsupportedEncodingException e1) { 
     // TODO Auto-generated catch block 
     e1.printStackTrace(); 
    }