【发布时间】:2018-05-08 01:38:27
【问题描述】:
我正在使用 nodejs 作为后端语言来制作登录/注册 Android 应用程序 UI。发生的事情是,每当我单击注册按钮时,它会在查询数据库之前发送 json 响应(选择语句以检查用户名是否存在)导致 android 应用程序得到错误的响应,这导致我单击按钮两次以注册正确回复。例如,假设他们是数据库中名为 test 的用户名,我尝试使用用户名 test 进行注册,它会告诉我该用户名已被占用,如果擦除测试并输入,可以说“bob”,它不存在数据库它仍然说用户名已经被使用,即使它不是但是当我再次点击注册按钮时它会注册用户。我假设这是因为它是异步的(它在查询数据库时或之前发送 json 响应)。我怎样才能使这个同步或有其他方法来解决这个问题?
服务器文件:
var express = require('express');
var app = express();
var bodyParser = require('body-parser');
var mysql = require('mysql');
//connection
var con = mysql.createConnection({
host: "localhost",
user: "root",
password: "password",
database : "androidtest"
});
//use json
app.use(bodyParser.json());
app.use(bodyParser.urlencoded({ extended: false }));
//declare variables to hold values entered by the user
var name;
var username;
var password;
var age;
//boolean array to let frontend know what is going on
var response;
app.post('/', function(req, res) {
//retrieve variables
username = req.body.username;
name = req.body.name;
age = req.body.age;
password = req.body.password;
//query database
//check if username is taken
var select = "SELECT * FROM users WHERE username = ? LIMIT 1";
con.query(select, [username], function (err, results) {
if (err) throw err;
//if username is taken send json string 'exists' to android app
if(results.length) {
response = {"exists" : "true"};
//if username is available send string 'success' and add the user to the database
} else {
var add = "INSERT INTO users (name, username, age, password) VALUES (?, ?, ?, ?)";
response = {"success" : "true"};
con.query(add, [name, username, age, password]);
if (err) throw err;
console.log('row inserted');
}
});
//send json
res.json(response);
//prevents the functions from being executed more than once
res.end('/');
});
//listen on port 3000
app.listen(3000);
android 应用程序 java:
bRegister.setOnClickListener(new View.OnClickListener() {
@Override
public void onClick(View view) {
final String name = etName.getText().toString();
final String username = etUserName.getText().toString();
final String password = etPassword.getText().toString();
final int age = Integer.parseInt(etAge.getText().toString());
Response.Listener<String> responseListener = new Response.Listener<String>() {
@Override
public void onResponse(String response) {
try {
JSONObject jsonResponse = new JSONObject(response);
//Boolean exists = jsonResponse.getBoolean("exists");
String exists = jsonResponse.getString("exists");
if (exists.matches("true")) {
Toast toast = Toast.makeText(getApplicationContext(), "username already exists", Toast.LENGTH_SHORT);
toast.show();
}
} catch(Exception e) {
e.printStackTrace();
}
try {
JSONObject jsonResponse = new JSONObject(response);
//Boolean success = jsonResponse.getBoolean("success");
String success = jsonResponse.getString("success");
if (success.matches("true")) {
Intent intent = new Intent(RegisterActivity.this, LoginActivity.class);
RegisterActivity.this.startActivity(intent);
} else {
AlertDialog.Builder builder = new AlertDialog.Builder(RegisterActivity.this);
builder.setMessage("Register Failed")
.setNegativeButton("Retry", null)
.create()
.show();
}
} catch (JSONException e) {
e.printStackTrace();
}
}
};
RegisterRequest registerRequest = new RegisterRequest(name, username, age, password, responseListener);
RequestQueue queue = Volley.newRequestQueue(RegisterActivity.this);
queue.add(registerRequest);
}
});
}
【问题讨论】:
-
这里有几百个这样的帖子。将
res.json(...)放入数据库回调中,以便仅在您实际拥有数据时触发。而且,在res.json()之后不需要res.end()。 -
确实有效,我在问问题之前尝试过,但我使用 console.log 来显示检索到的变量,并且它们被显示不止一次,并且它们也被添加到数据库中超过一次,所以我添加了 res.end() 导致显示并仅添加到数据库一次。但是,现在我没有使用 console.log(),而且它们似乎没有被多次添加到数据库中
-
我的回答对您有帮助吗?现在可以用了吗?
-
是的,现在可以了
标签: java android node.js rest express