Question

如果同样的问题已经发布在某些地方，我表示道歉，但在搜索了将近5天之后，人们可以理解我是多么绝望，因为大多数答案似乎都没有给我一个解决方案。漫无边际。所以，我试图使用齐射将json发布到wcf restful web服务。以下是我的代码片段

// Android / Volley

public void setJsonDataToWebservices(String url){
        JSONObject params = new JSONObject();
        try {
            params.put("FullName","nathalue Gustava");
            params.put("Id","123");
            params.put("Password","123456789");
        } catch (JSONException e) {
            e.printStackTrace();
        }
        // Create a future request object

        // Create a json request object
        JsonObjectRequest jsonObjectRequest = new JsonObjectRequest(Request.Method.POST, url, params,
                new Response.Listener<JSONObject>() {
                    @Override
                    public void onResponse(JSONObject jsonObject) {
                     Toast.makeText(getActivity(), "Success", Toast.LENGTH_SHORT).show();
                    }
                }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError volleyError) {
                Toast.makeText(getActivity(), "Failed", Toast.LENGTH_SHORT).show();
            }
        }){
            @Override
            public Map<String, String> getHeaders() throws AuthFailureError {
                HashMap<String, String> headers = new HashMap<String, String>();
                headers.put("Content-Type", "application/json; charset=utf-8");
                return headers;
            }
        };
        VolleySingleton.getIntance().addToRequestQueue(jsonObjectRequest);

    }

这是我的服务界面

[OperationContract]
        [WebInvoke(Method = "POST", RequestFormat = WebMessageFormat.Json, BodyStyle = WebMessageBodyStyle.WrappedRequest, UriTemplate = "simplePostExample")]
    string simplePostExample(Stream jsonStream);

......这是一个反序列化流并将其返回一个字符串的方法

public string simplePostExample(Stream jsonStream)
        {
            StreamReader reader = null;
            string text = string.Empty;
            try
            {
                reader = new StreamReader(jsonStream);
                text = reader.ReadToEnd();
            }
            catch (Exception)
            {

                throw;
            }

            return text != null ? text : "String is empty";
        }

以下是错误日志处理异常。异常详细信息：System.InvalidOperationException：操作的传入消息＆＃39; simplePostExample＆＃39; （合同＆＃39; IDigitizingDataRestfulWebService＆＃39; with namespace＆＃39; http://tempuri.org/＆＃39;）包含无法识别的http正文格式值＆＃39; Json＆＃39;。预期的正文格式值是＆＃39; Raw＆＃39;。这可能是因为尚未在绑定上配置WebContentTypeMapper。有关更多详细信息，请参阅WebContentTypeMapper的文档。在System.ServiceModel.Dispatcher.HttpStreamFormatter.GetStreamFromMessage（消息消息，布尔isRequest） at System.ServiceModel.Dispatcher.HttpStreamFormatter.DeserializeRequest（Message message，Object [] parameters）在System.ServiceModel.Dispatcher.UriTemplateDispatchFormatter.DeserializeRequest（消息消息，Object []参数）在System.ServiceModel.Dispatcher.CompositeDispatchFormatter.DeserializeRequest（消息消息，Object []参数）在System.ServiceModel.Dispatcher.DispatchOperationRuntime.DeserializeInputs（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.DispatchOperationRuntime.InvokeBegin（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage5（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage41（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage4（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage31（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage3（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage2（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage11（MessageRpc＆amp; rpc）在System.ServiceModel.Dispatcher.ImmutableDispatchRuntime.ProcessMessage1（MessageRpc＆amp; rpc）

现在，亲爱的人们，请告诉我哪里出错了。

使用Android Volley Library将JSON发布到WCF Restful Service

0 个答案: