我在市场上从我的应用程序获得用户报告,交付以下异常:

java.lang.IllegalStateException: Can not perform this action after onSaveInstanceState
at android.app.FragmentManagerImpl.checkStateLoss(FragmentManager.java:1109)
at android.app.FragmentManagerImpl.popBackStackImmediate(FragmentManager.java:399)
at android.app.Activity.onBackPressed(Activity.java:2066)
at android.app.Activity.onKeyUp(Activity.java:2044)
at android.view.KeyEvent.dispatch(KeyEvent.java:2529)
at android.app.Activity.dispatchKeyEvent(Activity.java:2274)
at com.android.internal.policy.impl.PhoneWindow$DecorView.dispatchKeyEvent(PhoneWindow.java:1803)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at com.android.internal.policy.impl.PhoneWindow$DecorView.superDispatchKeyEvent(PhoneWindow.java:1855)
at com.android.internal.policy.impl.PhoneWindow.superDispatchKeyEvent(PhoneWindow.java:1277)
at android.app.Activity.dispatchKeyEvent(Activity.java:2269)
at com.android.internal.policy.impl.PhoneWindow$DecorView.dispatchKeyEvent(PhoneWindow.java:1803)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.widget.TabHost.dispatchKeyEvent(TabHost.java:297)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at android.view.ViewGroup.dispatchKeyEvent(ViewGroup.java:1112)
at com.android.internal.policy.impl.PhoneWindow$DecorView.superDispatchKeyEvent(PhoneWindow.java:1855)
at com.android.internal.policy.impl.PhoneWindow.superDispatchKeyEvent(PhoneWindow.java:1277)
at android.app.Activity.dispatchKeyEvent(Activity.java:2269)
at com.android.internal.policy.impl.PhoneWindow$DecorView.dispatchKeyEvent(PhoneWindow.java:1803)
at android.view.ViewRoot.deliverKeyEventPostIme(ViewRoot.java:2880)
at android.view.ViewRoot.handleFinishedEvent(ViewRoot.java:2853)
at android.view.ViewRoot.handleMessage(ViewRoot.java:2028)
at android.os.Handler.dispatchMessage(Handler.java:99)
at android.os.Looper.loop(Looper.java:132)
at android.app.ActivityThread.main(ActivityThread.java:4028)
at java.lang.reflect.Method.invokeNative(Native Method)
at java.lang.reflect.Method.invoke(Method.java:491)
at com.android.internal.os.ZygoteInit$MethodAndArgsCaller.run(ZygoteInit.java:844)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:602)
at dalvik.system.NativeStart.main(Native Method)

显然这与FragmentManager有关,而我并不使用它。堆栈跟踪没有显示任何我自己的类,所以我不知道这个异常发生在哪里以及如何防止它。

为了记录:我有一个tabhost,在每个选项卡中都有一个在活动之间切换的ActivityGroup。


当前回答

为了绕过这个问题,我们可以使用谷歌I/O 2018中引入的导航体系结构组件。 导航架构组件简化了Android应用程序中导航的实现。

其他回答

异常在这里抛出(在FragmentActivity中):

@Override
public void onBackPressed() {
    if (!mFragments.getSupportFragmentManager().popBackStackImmediate()) {
        super.onBackPressed();
    }
}

在FragmentManager.popBackStatckImmediate()中,FragmentManager.checkStateLoss()首先被调用。这就是IllegalStateException的原因。参见下面的实现:

private void checkStateLoss() {
    if (mStateSaved) { // Boom!
        throw new IllegalStateException(
                "Can not perform this action after onSaveInstanceState");
    }
    if (mNoTransactionsBecause != null) {
        throw new IllegalStateException(
                "Can not perform this action inside of " + mNoTransactionsBecause);
    }
}

我通过使用一个标志来标记Activity的当前状态来解决这个问题。以下是我的解决方案:

public class MainActivity extends AppCompatActivity {
    /**
     * A flag that marks whether current Activity has saved its instance state
     */
    private boolean mHasSaveInstanceState;

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

    @Override
    protected void onSaveInstanceState(Bundle outState) {
        mHasSaveInstanceState = true;
        super.onSaveInstanceState(outState);
    }

    @Override
    protected void onResume() {
        super.onResume();
        mHasSaveInstanceState = false;
    }

    @Override
    public void onBackPressed() {
        if (!mHasSaveInstanceState) {
            // avoid FragmentManager.checkStateLoss()'s throwing IllegalStateException
            super.onBackPressed();
        }
    }
}

我最终创建了一个基本片段,并使我的应用程序中的所有片段扩展它

public class BaseFragment extends Fragment {

    private boolean mStateSaved;

    @CallSuper
    @Override
    public void onSaveInstanceState(Bundle outState) {
        mStateSaved = true;
        super.onSaveInstanceState(outState);
    }

    /**
     * Version of {@link #show(FragmentManager, String)} that no-ops when an IllegalStateException
     * would otherwise occur.
     */
    public void showAllowingStateLoss(FragmentManager manager, String tag) {
        // API 26 added this convenient method
        if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.O) {
            if (manager.isStateSaved()) {
                return;
            }
        }

        if (mStateSaved) {
            return;
        }

        show(manager, tag);
    }
}

然后,当我试图显示一个片段时,我使用showAllowingStateLoss而不是show

是这样的:

MyFragment.newInstance()
.showAllowingStateLoss(getFragmentManager(), MY_FRAGMENT.TAG);

我从这个PR上找到了这个解决方案:https://github.com/googlesamples/easypermissions/pull/170/files

在我的情况下,我在一个叫做onActivityResult的覆盖方法中得到了这个错误。在挖掘之后,我才发现也许我之前需要打电话给“超级”。 我加进去了,效果很好

override fun onActivityResult(requestCode: Int, resultCode: Int, data: Intent?) {
    super.onActivityResult(requestCode, resultCode, data); //<--- THIS IS THE SUPPER CALL
    if (resultCode == Activity.RESULT_OK && requestCode == 0) {
        mostrarFragment(FiltroFragment.newInstance())
    }

}

也许你只需要在代码之前的任何重写上添加一个“super”。

有许多与类似错误消息相关的问题。检查这个特定堆栈跟踪的第二行。这个异常与FragmentManagerImpl.popBackStackImmediate调用相关。

这个方法调用,就像popBackStack一样,如果会话状态已经被保存,它总是会以IllegalStateException失败。检查来源。您无法阻止抛出此异常。

删除对super的调用。onSaveInstanceState没有帮助。 使用commitAllowingStateLoss创建Fragment是没有帮助的。

以下是我对这个问题的观察:

有一个带有提交按钮的表单。 当单击按钮时,将创建一个对话框并启动异步进程。 用户在过程结束之前单击home键—调用onSaveInstanceState。 进程完成后,将执行回调并尝试popBackStackImmediate。 抛出IllegalStateException。

下面是我解决问题的方法:

因为在回调中不可能避免IllegalStateException,所以捕获并忽略它。

try {
    activity.getSupportFragmentManager().popBackStackImmediate(name);
} catch (IllegalStateException ignored) {
    // There's no way to avoid getting this if saveInstanceState has already been called.
}

这足以阻止应用程序崩溃。但现在用户将恢复应用程序,并看到他们认为他们按下的按钮根本没有按下(他们认为)。表单片段仍然显示!

要修复此问题,在创建对话框时,使某些状态表示进程已启动。

progressDialog.show(fragmentManager, TAG);
submitPressed = true;

并将这个状态保存在bundle中。

@Override
public void onSaveInstanceState(Bundle outState) {
    ...
    outState.putBoolean(SUBMIT_PRESSED, submitPressed);
}

不要忘记在onViewCreated中再次加载它

然后,在恢复时,如果之前尝试提交,则回滚这些片段。这可以防止用户返回到一个似乎未提交的表单。

@Override
public void onResume() {
    super.onResume();
    if (submitPressed) {
        // no need to try-catch this, because we are not in a callback
        activity.getSupportFragmentManager().popBackStackImmediate(name);
        submitPressed = false;
    }
}

另一个可能的解决方法,我不确定是否在所有情况下都有帮助(来源在这里):

@Override
protected void onSaveInstanceState(Bundle outState) {
    super.onSaveInstanceState(outState);
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT) {
        final View rootView = findViewById(android.R.id.content);
        if (rootView != null) {
            rootView.cancelPendingInputEvents();
        }
    }
}