安卓错误 : "commit already called" when remove a fragment

Question

我想实现一个有 fragment 的 Activity !当我点击 Fragment1 时， Fragment2 被调用，当我点击 Fragment2 时， Fragment2 应该从屏幕上删除!我通过在其 onCreateView 中调用 Fragment2 的 LinearLayout 的 setOnclickListener 来实现它，在我的 onclicklistener 上，我调用了

transaction.remove(myFragment);

transaction.commit();

但在那之后我遇到了这个错误:commit already called我该如何解决这个错误，这是我的代码:这是我的 fragment 类

public class ArticleFragment extends Fragment {
final static String ARG_POSITION = "position";
int mCurrentPosition = -1;
private static LinearLayout l;
private android.support.v4.app.FragmentTransaction transaction;
private Fragment newFragment;

public void setArticleFragment(android.support.v4.app.FragmentTransaction transaction , Fragment newFragment) {
    this.transaction = transaction;
    this.newFragment = newFragment;
}

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, 
    Bundle savedInstanceState) {
    if (savedInstanceState != null) {
        mCurrentPosition = savedInstanceState.getInt(ARG_POSITION);
    }
    View v = inflater.inflate(R.layout.article_view, container , false);
    l = (LinearLayout) v.findViewById(R.id.transparentArea);
    if(l == null)
    l.setOnClickListener(new OnClickListener() {

        @Override
        public void onClick(View arg0) {
            // TODO Auto-generated method stub
            transaction.remove(newFragment);
            transaction.commit();
        }
    });
    return v;
}
}

和我的 mainAcitivity 类

public class MainActivity extends FragmentActivity implements
    HeadlinesFragment.OnHeadlineSelectedListener{

HeadlinesFragment firstFragment;

@Override
public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.news_articles);
    if (findViewById(R.id.fragment_container) != null) {
        if (savedInstanceState != null) {
            return;
        }
        firstFragment = new HeadlinesFragment();
        firstFragment.setArguments(getIntent().getExtras());
        getSupportFragmentManager().beginTransaction()
                .add(R.id.fragment_container, firstFragment).commit();
    }
}

public void onArticleSelected(int position) {
    ArticleFragment articleFrag = (ArticleFragment) getSupportFragmentManager()
            .findFragmentById(R.id.article_fragment);

    if (articleFrag != null) {
        articleFrag.updateArticleView(position);

    } else {
        final ArticleFragment newFragment = new ArticleFragment();
        Bundle args = new Bundle();
        args.putInt(ArticleFragment.ARG_POSITION, position);
        newFragment.setArguments(args);

        final FragmentTransaction transaction = getSupportFragmentManager()
                .beginTransaction();
        newFragment.setArticleFragment(transaction, newFragment);

        transaction.setCustomAnimations(R.anim.slidein, R.anim.slideout);
        transaction.add(R.id.fragment_container, newFragment);
        transaction.commit();

    }
}

Answer 1

不是重用传入的事务，而是创建一个新的 FragmentTransaction 实例来删除 Fragment2。

更简单的方法是将第一个 fragment 事务添加到 fragment 返回堆栈(例如 addToBackStack(null))，然后在 Fragment2 中，只需使用 FragmentManager 弹出返回堆栈 popBackStack()。