使用Volley POST传递参数

Question

我能够使用Postman和这些参数调用HTTP端点:

{
    "name":"Val",
    "subject":"Test"
}

但是我无法通过Android对Volley做同样的事情:这里尝试使用JSONRequest:

HashMap<String, String> params2 = new HashMap<String, String>();
        params.put("name", "Val");
        params.put("subject", "Test Subject");

        JsonObjectRequest jsObjRequest = new JsonObjectRequest
                (Request.Method.POST, Constants.CLOUD_URL, new JSONObject(params2), new Response.Listener<JSONObject>() {

                    @Override
                    public void onResponse(JSONObject response) {
                        mView.showMessage("Response: " + response.toString());
                    }
                }, new Response.ErrorListener() {

                    @Override
                    public void onErrorResponse(VolleyError error) {
                        // TODO Auto-generated method stub
                        mView.showMessage(error.getMessage());

                    }
                });

        // Access the RequestQueue through your singleton class.
        VolleySingleton.getInstance(mContext).addToRequestQueue(jsObjRequest);

这里是尝试StringRequest

private void postMessage(Context context, final String name, final String subject ){

        RequestQueue queue = Volley.newRequestQueue(context);
        StringRequest sr = new StringRequest(Request.Method.POST, Constants.CLOUD_URL, new Response.Listener<String>() {
            @Override
            public void onResponse(String response) {
                mView.showMessage(response);
            }
        }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {

            }
        }){
            @Override
            protected Map<String,String> getParams(){
                Map<String,String> params = new HashMap<String, String>();
                params.put("name", name);
                params.put("subject", subject);

                return params;
            }

            @Override
            public Map<String, String> getHeaders() throws AuthFailureError {
                Map<String,String> params = new HashMap<String, String>();
                params.put("Content-Type","application/x-www-form-urlencoded");
                return params;
            }
        };
        queue.add(sr);
    }

当我使用JSONRequest时,调用POST但没有参数传递,当我使用StringRequest时,我得到下面的错误？如何将JSON数据传递给Volley电话？

E/Volley: [13053] BasicNetwork.performRequest: Unexpected response code 400 for 

这是处理请求的服务器代码

public static async Task<HttpResponseMessage> Run(HttpRequestMessage req, TraceWriter log)
{
    var helloRequest = await req.Content.ReadAsAsync<HelloRequest>();

    var name = helloRequest?.Name ?? "world";    
    var responseMessage = $"Hello {personToGreet}!";


    log.Info($"Message: {responseMessage}");


    return req.CreateResponse(HttpStatusCode.OK, $"All went well.");
}

public class HelloRequest
{
    public string Name { get; set; }
    public string Subject { get; set; }
}

Answer 1

服务器代码期望JSON对象返回字符串或更确切地说是Json字符串.

JsonObjectRequest

JSONRequest在请求主体中发送JSON对象,并期望响应中有JSON对象.由于服务器返回一个字符串,它最终会抛出ParseError

StringRequest

StringRequest发送一个具有正文类型的请求,x-www-form-urlencoded但由于服务器需要一个JSON对象.您最终得到400 Bad Request

解决方案

解决方案是将字符串请求中的内容类型更改为JSON,并在正文中传递JSON对象.因为它已经期待你回复的字符串,所以你很好.代码应如下.

StringRequest sr = new StringRequest(Request.Method.POST, Constants.CLOUD_URL, new Response.Listener<String>() {
    @Override
    public void onResponse(String response) {
        mView.showMessage(response);
    }
}, new Response.ErrorListener() {
    @Override
    public void onErrorResponse(VolleyError error) {
        mView.showMessage(error.getMessage());
    }
}) {
    @Override
    public byte[] getBody() throws AuthFailureError {
        HashMap<String, String> params2 = new HashMap<String, String>();
        params2.put("name", "Val");
        params2.put("subject", "Test Subject");
        return new JSONObject(params2).toString().getBytes();
    }

    @Override
    public String getBodyContentType() {
        return "application/json";
    }
};

此外,服务器代码中还有一个错误

var responseMessage = $"Hello {personToGreet}!";

应该

var responseMessage = $"Hello {name}!";