Question

我在Android和Android中相对较新宁静的API编程，所以我可能会犯一些愚蠢的错误。但是我从JsonObject的回复中面对这个：

对于值\＆＃34;登录\＆＃34;，转换为ObjectId失败在路径\＆＃34; _id \＆＃34; for model \＆＃34; User \＆＃34;＆＃34;，＆＃34; name＆＃34;：＆＃34; CastError＆＃34;，＆＃34; stringValue＆＃34;：＆＃34; \ ＆＃34;登录\＆＃34;＆＃34;＆＃34;种类＆＃34;：＆＃34;的ObjectId＆＃34;＆＃34;值＆＃34;：＆＃34;登录＆＃34; ＆＃34;路径＆＃34;：＆＃34; _id＆＃34;}

我在Postman上测试了我的Rest API，一切都恢复正常。 Restful API是用Node.js和Express编写的。

Rest API假设返回状态，用户ID和消息，以便让应用知道用户是否使用正确的用户名和密码登录。登录工作正常，它返回200，允许应用程序登录并使用会话登录。但是，我无法从响应JsonObject获取任何内容以获取用户ID，以显示用户信息的下一步。

（目标Android API 23）

代码：

Restful.Java

public class RestClient {

    private static final String BASE_URL = "http://10.0.2.2:8000/api/";
    private static AsyncHttpClient client = new AsyncHttpClient();

    public static void get(String url, RequestParams params, AsyncHttpResponseHandler responseHandler){
        client.get(getAbsoluteUrl(url), params, responseHandler);
    }

    public static void post(String url, RequestParams params, AsyncHttpResponseHandler responseHandler){
        client.post(getAbsoluteUrl(url), params, responseHandler);
    }

    private static String getAbsoluteUrl(String relativeUrl){
        Log.d("URL: ", BASE_URL+relativeUrl);
        return BASE_URL + relativeUrl;
    }
}

LoginActivity.Java（仅限登录部分）

public void checkLoginDB(final RequestParams params, final String username){
        prgDialog.show();     

        RestClient.post("user/login", params, new JsonHttpResponseHandler(){
            @Override
            public void onSuccess(int statusCode, Header[] headers, JSONArray response) {
                Log.d("StatusCode: ", "Code "+ statusCode);
                try {
                    if (statusCode == 200) {
                        Toast.makeText(getApplicationContext(), "Logged In!", Toast.LENGTH_LONG).show();
                        session.createLoginSession(username);
                        Log.d("Log: ", "what " + response);
                        navigatetoHomeActivity();
                    }
                } catch (Exception e){
                    e.printStackTrace();
                }
            }

            @Override
            public void onFailure(int statusCode, Header[] headers, String responseString, Throwable throwable) {
                Log.d("StatusCode: ", "Code "+ statusCode);
                prgDialog.hide();
                Toast.makeText(getApplicationContext(), "Failed!", Toast.LENGTH_LONG).show();
            }
        });
    }

Restful API的Server.js

  router.post('/user/login', function(req, res){
    var username = req.body.username;
    var password = req.body.password;
    if(username.length > 0 && password.length > 3){
      User.findOne({username: username, password: password}, function(err, user){
        if(err)
          res.json({statusCode: 0, message: "login error: " + err});
        if(!user)
          res.json({statusCode: 0, message: "Not Found"});

        res.json({statusCode: 200, userid: user._id, message: "Login Sucessful"});
      })
    } else {
      res.json({statusCode: 0, message: "Invalid Fields"});
    }
  });

  app.use('/api', router)

User.js（已定义架构）

var mongoose = require('mongoose');
var Schema = mongoose.Schema;

var UserSchema = new Schema({
  username: String,
  password: String,
  firstname: String,
  lastname: String,
  agentID: Number,
  email: String,
  phone: Number
});

module.exports = mongoose.model('User', UserSchema);

Answer 1

每当您想要登录用户并希望返回时使用post（）方法的用户详细信息。在您的代码中，第二种方法不会发布，因此请将client.get()更改为client.post()

喜欢这个public static void post(String url, RequestParams params, AsyncHttpResponseHandler responseHandler){ client.post(getAbsoluteUrl(url), params, responseHandler); }

Android Loopj的Async JsonObject响应被强制转换为Objectid失败

1 个答案: