java - Android PHP POST 登录

Question

我尝试在其他帖子中寻找答案并避免重复发布；然而，许多人使用的是已弃用的方法。我正在尝试创建一个登录 Activity ，它将用户详细信息发布到 php 并接收 jsonEncoded 状态“已授权”或“用户详细信息不正确”响应；我已经测试了 PHP，它可以工作。当我单击登录按钮时，它将执行以下操作:

@Override
public void onClick(View v) {
    switch (v.getId()) {
        case R.id.btnLogin:

            String username = Username.getText().toString();
            String password = Password.getText().toString();

            User user = new User(username, password);

            authenticate(user);

            userLocalStore.setUserLoggedIn(true);
            userLocalStore.storeUserData(user);

            break;
    }
}

它包含了User函数，这让我更容易传递信息。

String username, password;

public User (String username, String password) {
    this.username = username;
    this.password = password;
}

构造用户后，登录将调用身份验证函数:

private void authenticate(User user) {

    ServerRequests serverRequests = new ServerRequests(this);
    serverRequests.fetchUserDataAsyncTask(user, returnedUser);{

        if (returnedUser == null){
            showErrorMessage();
        } else {
            logUserIn();
        }
    };
}

用户通过身份验证后，它将登录用户并启动新 Activity :

private void logUserIn() {
    userLocalStore.setUserLoggedIn(true);
    startActivity(new Intent(this, Map.class));
}

我遇到的问题是，它似乎没有调用 php，而且我不确定哪里出了问题；我尝试过将其他人所做的事情拼凑在一起，但它不起作用。如果有任何帮助，我将不胜感激。

import android.app.ProgressDialog;
import android.content.Context;
import android.os.AsyncTask;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.net.HttpURLConnection;
import java.net.MalformedURLException;
import java.net.ProtocolException;
import java.net.URL;

public class ServerRequests {

    URL url;
    HttpURLConnection conn;
    ProgressDialog progressDialog;

    public ServerRequests(Context context) {
        progressDialog = new ProgressDialog(context);
        progressDialog.setCancelable(false);
        progressDialog.setTitle("Authenticating...");
        progressDialog.setMessage("Please wait...");
    }

    public void fetchUserDataAsyncTask(User user, String returnedUser) {
        progressDialog.show();
        new fetchUserDataAsyncTask(user, returnedUser).execute();
    }

    public class fetchUserDataAsyncTask extends AsyncTask<Void, Void, String> {
        User user;
        String returnedUser;


        public fetchUserDataAsyncTask(User user, String returnedUser) {
            this.user = user;
            this.returnedUser = returnedUser;
        }

        @Override
        protected String doInBackground(Void... params) {

            try {

                url = new URL("www.mysite.com/test.php");

                conn = (HttpURLConnection) url.openConnection();
                String param = "username="+user.username+"&password="+user.password;
                conn.setRequestMethod("POST");
                conn.setDoInput(true);
                conn.setDoOutput(true);
                conn.setRequestProperty("Content-Type", "application/x-www-form-urlencoded");
                conn.setRequestProperty("charset", "UTF-8");
                conn.connect();

                OutputStreamWriter out = new OutputStreamWriter(conn.getOutputStream());
                out.write(param);

               //No code to receive response yet, want to get the POST working first. 

            }
            catch (MalformedURLException e) {
                e.printStackTrace();
            } catch (ProtocolException e) {
                e.printStackTrace();
            } catch (IOException e) {
                e.printStackTrace();
            }
            catch (Exception e) {

            } finally {
                progressDialog.dismiss();
            }

            return returnedUser;
        }
    }
}

Answer 1

试试这个..

conn = (HttpURLConnection) url.openConnection();
conn.setRequestMethod("POST");
conn.setDoOutput(true);
String param = "username="+user.username+"&password="+user.password;
OutputStream out = conn.getOutputStream();
DataOutputStream dataOut = new DataOutputStream(out);
dataOut.writeBytes(param.trim());
InputStream in = conn.getInputStream();
BufferedReader reader = new BufferedReader(new InputStreamReader(in));
String line;
StringBuilder builder = new StringBuilder();
while((line = reader.readLine()) != null) {
    builder.append(line);
}
reader.close();
System.out.println(builder.toString());

在StringBuilder中，您将获得服务器返回的输出。

它将把你的参数提交到 php 页面..