Android Volley:BasicNetwork.performRequest:意外的响应代码404

时间:2017-04-06 17:48:31

标签: java android json android-volley

我想从数据库中获取用户的信息并显示它,但我一直收到此错误:

  

BasicNetwork.performRequest:意外的响应代码404

这是我的javacode:

加载方法:

public void loadUserProfile() {

    Response.Listener<String> responseListener = new Response.Listener<String>() {
        @Override
        public void onResponse(String response) {
            try {
                JSONObject jsonResponse = new JSONObject(response);
                idTV.setText(jsonResponse.getString("_id"));
                nameTV.setText(jsonResponse.getString("name"));
                firstNameTV.setText(jsonResponse.getString("firstName"));
                emailTV.setText(jsonResponse.getString("email"));
                sexTV.setText(jsonResponse.getString("sex"));
                yearTV.setText(jsonResponse.getString("year"));
                cursusTV.setText(jsonResponse.getString("cursus"));
            } catch (JSONException e) {
                onLoadProfileFail();
            }
        }
    };
    ProfileRequest profileRequest = new ProfileRequest(user_id, responseListener);
    RequestQueue queue = Volley.newRequestQueue(ProfileActivity.this);
    queue.add(profileRequest);
}

请求类

class ProfileRequest extends StringRequest {
    private static final String USER_PROFILE_REQUEST = "http://[ipaddress]:8080/users/:id";
    private Map<String, String> params;

    public ProfileRequest(String id, Response.Listener<String> listener) {
        super(Method.POST, USER_PROFILE_REQUEST, listener, null);
        params = new HashMap<>();
        params.put("id", id);
    }

    @Override
    public Map<String, String> getParams() {
        return params;
    }
}

app.js

router.get('/users/:id', function(req, res) {
    User.find({ _id: "ie00847" }, function(err, user) {
      if (err)
          res.send(err);
      res.json(user);
    });
});

2 个答案:

答案 0 :(得分:0)

由于您使用不可用的方法(POST)访问路由,因此出现404错误,因为在app.js文件中您为get请求而不是POST请求定义了一个函数,因此您需要将Android代码更新为GET请求。如下所示,

class ProfileRequest extends StringRequest {
    private static final String USER_PROFILE_REQUEST = "http://[ipaddress]:8080/users";
    private Map<String, String> params;

    public ProfileRequest(String id, Response.Listener<String> listener) {
        super(Method.GET, USER_PROFILE_REQUEST + "/" + id, listener, null);
    }
}

另请注意,由于您不再发送POST请求,因此无需覆盖getParams()方法。此外,除了通过对URL中的值进行硬编码而不是通过Map对象进行硬编码之外,您无法向GET请求添加参数。

希望这会有所帮助:)

答案 1 :(得分:0)

您可以使用此代码来解决

public void onErrorResponse(VolleyError error) {

                    NetworkResponse response = error.networkResponse;
                    if (response != null && response.statusCode == 404) {
                        try {
                            String res = new String(response.data,
                                    HttpHeaderParser.parseCharset(response.headers, "utf-8"));
                            // Now you can use any deserializer to make sense of data
                            JSONObject obj = new JSONObject(res);
                            //use this json as you want
                        } catch (UnsupportedEncodingException e1) {
                            // Couldn't properly decode data to string
                            e1.printStackTrace();
                        } catch (JSONException e2) {
                            // returned data is not JSONObject?
                            e2.printStackTrace();
                        }
                    }
                }