使用 volley 自定义发布请求并错误接收

Question

这是我的CustomRequest 代码。我正在使用 Map 传递参数以发布。

Map<String, String> params = new HashMap<String, String>();
    params.put("name", username);
    params.put("buildingname", userbuildingname);
    params.put("area", userarea);
    params.put("city", usercity);
    params.put("mobileno", usermobileno);
    params.put("email", useremail);
    params.put("confirmpassword", userconfirmpassword);
    params.put("method", method);
    params.put("format", format);


    CustomRequest jsonObjReq = new CustomRequest(Request.Method.POST, url, params, new Response.Listener<JSONObject>() {
                @Override
                public void onResponse(JSONObject response) {
                    JSONObject resp = response;
                    abc = resp.toString();
                }
            }, new Response.ErrorListener() {
                 @Override
                 public void onErrorResponse(VolleyError error) {
                     abc = error.getMessage();

                     Log.d(TAG, "Error: " + error.getMessage());
                 }
            });
    AppController.getInstance().addToRequestQueue(jsonObjReq);

我创建了自定义请求帮助程序类。使用这个link。但它也不适用于我。

import java.io.UnsupportedEncodingException;
import java.util.Map;

import org.json.JSONException;
import org.json.JSONObject;

import com.android.volley.NetworkResponse;
import com.android.volley.ParseError;
import com.android.volley.Request;
import com.android.volley.Response;
import com.android.volley.Response.ErrorListener;
import com.android.volley.Response.Listener;
import com.android.volley.toolbox.HttpHeaderParser;

public class CustomRequest extends Request<JSONObject>{

private Listener<JSONObject> listener;
private Map<String, String> params;

public CustomRequest(String url, Map<String, String> params,
                     Listener<JSONObject> reponseListener, ErrorListener errorListener) {
    super(Method.GET, url, errorListener);
    this.listener = reponseListener;
    this.params = params;
}

public CustomRequest(int method, String url, Map<String, String> params,
                     Listener<JSONObject> reponseListener, ErrorListener errorListener) {
    super(method, url, errorListener);
    this.listener = reponseListener;
    this.params = params;
}

@Override
protected Map<String, String> getParams() throws com.android.volley.AuthFailureError {
    return params;
};

@Override
protected void deliverResponse(JSONObject response) {
    listener.onResponse(response);
}

@Override
protected Response<JSONObject> parseNetworkResponse(NetworkResponse response) {
    try {
        String jsonString = new String(response.data,
                HttpHeaderParser.parseCharset(response.headers));
        return Response.success(new JSONObject(jsonString),
                HttpHeaderParser.parseCacheHeaders(response));
    } catch (UnsupportedEncodingException e) {
        return Response.error(new ParseError(e));
    } catch (JSONException je) {
        return Response.error(new ParseError(je));
    }
}

}

PHP- 我的 API 代码

else if(strcasecmp($_POST['method'],'userRegistration') == 0){
    $name = $_POST['name'];
    $buildingname= $_POST['buildingname'];
    $area= $_POST['area'];
    $city= $_POST['city'];
    $mobileno= $_POST['mobileno'];
    $email= $_POST['email'];
    $password= $_POST['confirmpassword'];
    deliver_response($_POST['format'],$response,false);
}

响应是这样传递的

$json_response = json_encode($api_response);

    // Deliver formatted data
    echo $json_response;

但它给了我错误

com.android.volley.AuthFailureError

请帮忙

Answer 1

终于找到答案了。

JSONObject params = new JSONObject();
        try {
            params.put("method", method);
            params.put("format", format);
            params.put("name", username);
            params.put("buildingname", userbuildingname);
            params.put("area", userarea);
            params.put("city", usercity);
            params.put("mobileno", usermobileno);
            params.put("email", useremail);
            params.put("confirmpassword", userconfirmpassword);
        } catch (Exception e) {

        }
        JsonObjectRequest req = new JsonObjectRequest(Request.Method.POST, URL, params,
                new Response.Listener<JSONObject>() {
                    @Override
                    public void onResponse(JSONObject response) {
                        VolleyLog.v("Response:%n %s", response.toString());
                        SignUp signUp = new SignUp();
                        try {
                            returnResponse(response.getString("saveUsersDetailsResponse"));
                        } catch (JSONException e) {
                            e.printStackTrace();
                        }
                    }
                }, new Response.ErrorListener() {
            @Override
            public void onErrorResponse(VolleyError error) {
                VolleyLog.e("Error: ", error.getMessage());
            }
        }
        );
        AppController.getInstance().addToRequestQueue(req);

在 PHP 中，我们以 json 格式获取数据。现在我们必须将 json 数据转换为 php 可读。

$checkmethod = $_SERVER['REQUEST_METHOD'];
$var = file_get_contents("php://input");
$string = json_decode($var, TRUE);
$method = $string['method'];
$name = $string['name'];
$email = $string['email'];
$mobileno = $string['mobileno'];

我们通过解码json得到所有数据。