我有一个场景,在通过登录页面登录后,每个活动都会有一个注销按钮。

单击注销时,我将传递要注销的登录用户的会话id。有人能指导我如何让所有活动都可以使用会话id吗?

本案的任何替代方案


当前回答

您可以使用intent对象在活动之间发送数据。假设您有两个活动,即FirstActivity和SecondActivity。

内部第一活动:

使用意图:

i = new Intent(FirstActivity.this,SecondActivity.class);
i.putExtra("key", value);
startActivity(i)

内部第二活动

Bundle bundle= getIntent().getExtras();

现在,您可以使用不同的bundle类方法获取通过Key从FirstActivity传递的值。

例如。bundle.getString(“key”)、bundle.get-Double(“key“)、bundle.getInt(“key”)等。

其他回答

/*
 * If you are from transferring data from one class that doesn't
 * extend Activity, then you need to do something like this.
 */ 

public class abc {
    Context context;

    public abc(Context context) {
        this.context = context;
    }

    public void something() {
        context.startactivity(new Intent(context, anyone.class).putextra("key", value));
    }
}

这是我的最佳实践,当项目规模巨大且复杂时,它会帮助很大。

假设我有两个活动,LoginActivity和HomeActivity。我想将2个参数(用户名和密码)从LoginActivity传递到HomeActivity。

首先,我创建HomeIntent

public class HomeIntent extends Intent {

    private static final String ACTION_LOGIN = "action_login";
    private static final String ACTION_LOGOUT = "action_logout";

    private static final String ARG_USERNAME = "arg_username";
    private static final String ARG_PASSWORD = "arg_password";


    public HomeIntent(Context ctx, boolean isLogIn) {
        this(ctx);
        //set action type
        setAction(isLogIn ? ACTION_LOGIN : ACTION_LOGOUT);
    }

    public HomeIntent(Context ctx) {
        super(ctx, HomeActivity.class);
    }

    //This will be needed for receiving data
    public HomeIntent(Intent intent) {
        super(intent);
    }

    public void setData(String userName, String password) {
        putExtra(ARG_USERNAME, userName);
        putExtra(ARG_PASSWORD, password);
    }

    public String getUsername() {
        return getStringExtra(ARG_USERNAME);
    }

    public String getPassword() {
        return getStringExtra(ARG_PASSWORD);
    }

    //To separate the params is for which action, we should create action
    public boolean isActionLogIn() {
        return getAction().equals(ACTION_LOGIN);
    }

    public boolean isActionLogOut() {
        return getAction().equals(ACTION_LOGOUT);
    }
}

以下是我如何在LoginActivity中传递数据

public class LoginActivity extends AppCompatActivity {
    @Override
    protected void onCreate(@Nullable Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);

        String username = "phearum";
        String password = "pwd1133";
        final boolean isActionLogin = true;
        //Passing data to HomeActivity
        final HomeIntent homeIntent = new HomeIntent(this, isActionLogin);
        homeIntent.setData(username, password);
        startActivity(homeIntent);

    }
}

最后一步,这里是我如何在HomeActivity中接收数据

public class HomeActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_home);

        //This is how we receive the data from LoginActivity
        //Make sure you pass getIntent() to the HomeIntent constructor
        final HomeIntent homeIntent = new HomeIntent(getIntent());
        Log.d("HomeActivity", "Is action login?  " + homeIntent.isActionLogIn());
        Log.d("HomeActivity", "username: " + homeIntent.getUsername());
        Log.d("HomeActivity", "password: " + homeIntent.getPassword());
    }
}

完成!酷:)我只是想分享我的经验。如果你在做小项目,这应该不是大问题。但当你在大项目上工作时,当你想进行重构或修复bug时,这真的很痛苦。

在CurrentActivity.java中编写以下代码

Intent i = new Intent(CurrentActivity.this, SignOutActivity.class);
i.putExtra("SESSION_ID",sessionId);
startActivity(i);

按以下方式访问SignOutActivity.java中的SessionId

public void onCreate(Bundle savedInstanceState){
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_sign_out);
    Intent intent = getIntent();
    
    // check intent is null or not
    if(intent != null){
        String sessionId = intent.getStringExtra("SESSION_ID");
        Log.d("Session_id : " + sessionId);
    }
    else{
        Toast.makeText(SignOutActivity.this, "Intent is null", Toast.LENGTH_SHORT).show();
    }
}

来自活动

int n= 10;
Intent in = new Intent(From_Activity.this,To_Activity.class);
Bundle b1 = new Bundle();
b1.putInt("integerNumber",n);
in.putExtras(b1);
startActivity(in);

目标活动

Bundle b2 = getIntent().getExtras();
int m = 0;
if(b2 != null){
 m = b2.getInt("integerNumber");
}

Destination活动的定义如下:

public class DestinationActivity extends AppCompatActivity{

    public static Model model;
    public static void open(final Context ctx, Model model){
          DestinationActivity.model = model;
          ctx.startActivity(new Intent(ctx, DestinationActivity.class))
    }

    public void onCreate(/*Parameters*/){
           //Use model here
           model.getSomething();
    }
}

在第一个活动中,按如下方式开始第二个活动:

DestinationActivity.open(this,model);