Question

这是一个简单的登录活动，但当它必须切换活动时，应用程序崩溃。我尝试使用正常形式的pubic void onClick（）{...}但它不起作用。

Login.java

package com.example.corrado_mattia_danny.face_offbrains;

import android.app.Activity;
import android.content.Intent;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;
import android.widget.CheckedTextView;
import android.widget.EditText;
import android.widget.Toast;


public class Login extends Activity {

    private Player player;
    private Button login;
    private EditText Username;
    private EditText Password;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_login);
        login=(Button)findViewById(R.id.button_sign_in);
        Username=(EditText)findViewById(R.id.nickname);
        Password=(EditText)findViewById(R.id.password);
        String a = Username.getText().toString();
        String b = Password.getText().toString();
        player.inserisciCredenziali(a,b); //insert into a player class
                                          //username and password
        signIn();
    }

    public void signIn() {
        Intent intent = new Intent(this, Home.class);
        startActivity(intent);
    }
}

activity_login.xml

<RelativeLayout
    xmlns:android="http://schemas.android.com/apk/res/android"
    xmlns:tools="http://schemas.android.com/tools"
    android:layout_width="match_parent"
    android:layout_height="match_parent"
    android:paddingLeft="@dimen/activity_horizontal_margin"
    android:paddingRight="@dimen/activity_horizontal_margin"
    android:paddingTop="@dimen/activity_vertical_margin"
    tools:context="com.example.corrado_mattia_danny.face_offbrains.Login"
    android:background="@drawable/sfondo_custom" >


    <EditText
        android:layout_width="wrap_content"
        android:layout_height="35dp"
        android:inputType="textPersonName"
        android:hint="Nickname"
        android:ems="10"
        android:id="@+id/nickname"
        android:linksClickable="false"
        android:textColor="#ff000000"
        android:background="#ffd4d4d4"
        android:textStyle="normal|bold|italic"
        android:layout_marginTop="100dp"
        android:layout_alignParentTop="true"
        android:layout_centerHorizontal="true" />

    <EditText
        android:layout_width="wrap_content"
        android:layout_height="35dp"
        android:inputType="textPassword"
        android:ems="10"
        android:id="@+id/password"
        android:layout_below="@+id/nickname"
        android:layout_alignStart="@+id/nickname"
        android:layout_marginTop="30dp"
        android:hint="password"
        android:textColor="#ff000000"
        android:background="#ffd4d4d4"
        android:textStyle="normal|bold|italic" />

    <Button
        style="?android:attr/buttonStyleSmall"
        android:layout_width="50dp"
        android:layout_height="50dp"
        android:id="@+id/button_sign_in"
        android:background="#ffffeb00"
        android:layout_centerVertical="true"
        android:layout_toEndOf="@+id/nickname"
        android:layout_marginLeft="10dp"
        android:onClick="signIn"/>

</RelativeLayout>

Home.java

Home中的按钮无效。

package com.example.corrado_mattia_danny.face_offbrains;

import android.app.Activity;
import android.content.Intent;
import android.support.v7.app.ActionBarActivity;
import android.os.Bundle;
import android.view.Menu;
import android.view.MenuItem;
import android.view.View;
import android.widget.Button;


public class Home extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_home);
        Intent intent = getIntent();
        Button bottone_sfida_amico = (Button)findViewById(R.id.button_sfida_amico);
        Button bottone_avversario_casuale = (Button)findViewById(R.id.button_avversario_casuale);
    }
}

}

Answer 1

通过android:onClick属性设置方法时，方法signIn()必须包含参数View，这是事件的来源。因此，声明方法如下：

public void signIn(View view) {
    ...
}

Answer 2

可以是两件事： 1你在清单上声明了吗？

 <activity android:name=".SampleActivity" android:screenOrientation="landscape"
    android:label="SampleActivity"
    />

另一个原因必须是你的函数的声明，当你为布局调用它时它应该像这样（View v）作为param

 public void signIn(View v) {
    Intent intent = new Intent(this, Home.class);
    startActivity(intent);
}

应用程序崩溃使用意图切换活动

2 个答案: