Question

我正在使用需要登录的android应用程序，现在我想要成功登录我的应用程序后，登录按钮应该是不可见的，并且应该可以看到注销按钮，但是会出错。

我正在使用此功能：

login.setVisibility(View.GONE);    &    logout.setVisibility(View.VISIBLE);

Eclipse在logcat中给我错误，如：

java.lang.RuntimeException: Unable to start activity ComponentInfo{com.rahul.cheerfoolz/com.rahul.cheerfoolz.CheerfoolznativeActivity}: java.lang.NullPointerException.

Main.class

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);

      session = new SessionID();
      Button login = (Button) findViewById(R.id.home_btn_feature_login);
      Button logout = (Button) findViewById(R.id.home_btn_feature_logout);

         int userID = SessionID.getUserID();

         System.out.println("value of the userID into the main page:====>"+userID);
         //  when user login then here I got the userID value

        if ((userID) > 0) {

            login.setVisibility(View.GONE);  //when Execute this give me error
            logout.setVisibility(View.VISIBLE);  //when Execute this give me error

        } else {

            login.setVisibility(View.VISIBLE); //when Execute this give me error
            logout.setVisibility(View.GONE); //when Execute this give me error

        }

    setContentView(R.layout.main);
    setHeader_home("");
    }

main.xml中

   <LinearLayout
        android:layout_width="fill_parent"
        android:layout_height="wrap_content"
        android:layout_weight="2"
        android:orientation="horizontal" >

   <!-- Here Login Button  -->

        <Button
            android:id="@+id/home_btn_feature_login"
            style="@style/HomeButton"
            android:drawableTop="@drawable/login_button"
            android:onClick="onClickFeature"
            android:text="@string/title_feature_login" />

   <!-- Here Logout Button  -->

        <Button
            android:id="@+id/home_btn_feature_logout"
            style="@style/HomeButton"
            android:drawableTop="@drawable/logout_button"
            android:onClick="onClickFeature"
            android:text="@string/title_feature_logout" />

        <Button
            android:id="@+id/home_btn_feature2"
            style="@style/HomeButton"
            android:drawableTop="@drawable/register_button"
            android:onClick="onClickFeature"
            android:text="@string/title_feature_register" />
    </LinearLayout>

Answer 1

您必须在setContentView(R.layout.main)之后添加super.onCreate(savedInstanceState); 或before Accessing Any View from XML Layout File。然后你的Button从xml.and获取引用代码无法从main.xml获取Button引用.so NullPointerException。

protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.main);<<<<<<<<<<<<<<ADD THIS LINE


    session = new SessionID();
    Button login = (Button) findViewById(R.id.home_btn_feature_login);
    Button logout = (Button) findViewById(R.id.home_btn_feature_logout);

Answer 2

在setContentView(R.layout.main);

之后添加此super.onCreate(savedInstanceState);

按钮可见性/隐身问题。

2 个答案: