我正在尝试制作一个 ListFragment。我看了 Api Demo (FragmentLayout)。它适用于一个简单的示例，现在我想将它应用到我现有的项目中。
这是我的代码。我在 Api 演示中创建内部类（RecipeList 和 RecipeDetail）。
public class InfoActivity extends MenuActivity {
@Override
public void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.info_fragment_layout);
// ...
}
public static class RecipeList extends ListFragment {
private int mCurrentSelectedItemIndex = -1;
private boolean mIsTablet = false;
@Override
public void onActivityCreated(Bundle savedInstanceState) {
super.onActivityCreated(savedInstanceState);
accountData = new ArrayList<Account>();
new AccountSyncTask() {
@Override
public void onPostExecute(
final ArrayList<ArrayList<String>> result) {
// For each retrieved account
Bd.insert(retrievedAccount);
accountData.add(retrievedAccount);
}
accountListAdapter = new AccountListAdapter(
InfoActivity.this, R.layout.accountlist_detail,
accountData);
accountListAdapter = new AccountListAdapter(
activityContext, R.layout.accountlist_detail,
accountData);
setListAdapter(accountListAdapter);
}
}.execute(sessionName, null, "getAllObjectOnServer",
String.valueOf(nbRow));
if (savedInstanceState != null) {
mCurrentSelectedItemIndex = savedInstanceState.getInt(
"currentListIndex", -1);
}
// This is a tablet if this view exists
View recipeDetails = getActivity()
.findViewById(R.id.recipe_details);
mIsTablet = recipeDetails != null
&& recipeDetails.getVisibility() == View.VISIBLE;
if (mIsTablet) {
getListView().setChoiceMode(ListView.CHOICE_MODE_SINGLE);
}
if (mIsTablet && mCurrentSelectedItemIndex != -1) {
showRecipeDetails();
}
}
@Override
public void onListItemClick(ListView l, View v, int position, long id) {
mCurrentSelectedItemIndex = position;
showRecipeDetails();
}
private void showRecipeDetails() {
if (mIsTablet) {
// Set the list item as checked
getListView().setItemChecked(mCurrentSelectedItemIndex, true);
// Get the fragment instance
RecipeDetail details = (RecipeDetail) getFragmentManager()
.findFragmentById(R.id.recipe_details);
// Is the current visible recipe the same as the clicked? If so,
// there is no need to update
if (details == null
|| details.getRecipeIndex() != mCurrentSelectedItemIndex) {
details = RecipeDetail
.newInstance(mCurrentSelectedItemIndex);
FragmentTransaction ft = getFragmentManager()
.beginTransaction();
ft.replace(R.id.recipe_details, details);
ft.setTransition(FragmentTransaction.TRANSIT_FRAGMENT_FADE);
ft.commit();
}
} else {
}
}
@Override
public void onSaveInstanceState(Bundle outState) {
super.onSaveInstanceState(outState);
outState.putInt("currentListIndex", mCurrentSelectedItemIndex);
}
}
public static class RecipeDetail extends Fragment {
private int mRecipeIndex;
public static RecipeDetail newInstance(int recipeIndex) {
// Create a new fragment instance
RecipeDetail detail = new RecipeDetail();
// Set the recipe index
detail.setRecipeIndex(recipeIndex);
return detail;
}
@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
Bundle savedInstanceState) {
if (container == null) {
return null;
}
View v = inflater
.inflate(R.layout.recipe_details, container, false);
//..
return v;
}
public int getRecipeIndex() {
return mRecipeIndex;
}
public void setRecipeIndex(int index) {
mRecipeIndex = index;
}
}
我有一个自定义的 ArrayAdapter（我在 ListFragment 中的项目包含 4 个 textView 和一个可点击的 imageButton）。
帐户列表适配器：
public class AccountListAdapter extends ArrayAdapter<Account> {
private final Context context;
private final int layoutResourceId;
private final ArrayList<Account> data;
public AccountListAdapter(Context context, int layoutResourceId,
ArrayList<Account> data) {
super(context, layoutResourceId, data);
this.layoutResourceId = layoutResourceId;
this.context = context;
this.data = data;
}
@Override
public View getView(final int position, View convertView, ViewGroup parent) {
AccountHolder holder = null;
if (convertView == null) {
LayoutInflater inflater = ((Activity) context).getLayoutInflater();
convertView = inflater.inflate(layoutResourceId, parent, false);
holder = new AccountHolder();
convertView.setClickable(true);
convertView.setFocusable(true);
holder.txtName = (TextView) convertView.findViewById(R.id.nom);
holder.txtId = (TextView) convertView.findViewById(R.id.id);
convertView.setTag(holder);
} else {
holder = (AccountHolder) convertView.getTag();
}
convertView.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
Log.i("click", "index = " + position);
}
});
holder.txtName.setText(data.get(position).getName());
holder.txtId.setText(data.get(position).getId());
convertView.setBackgroundResource(R.drawable.list_selector);
ImageButton img = (ImageButton) convertView.findViewById(R.id.check);
img.setTag(position);
return convertView;
}
static class AccountHolder {
TextView txtName;
TextView txtId;
}
}
问题：当我点击 listFragment 的一个项目时，
public void onListItemClick(ListView l, View v, int position, long id) {
mCurrentSelectedItemIndex = position;
Log.i("click", "here";
showRecipeDetails();
}
未调用，但 AccountListAdapter 中定义的项目上的侦听器有效
convertView.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
Log.i("click", "index = " + position);
}
});
为什么从不调用 onListitemClick ？
另一个问题：在我的 ListFragment 的 onActivityCreated 函数中的另一个线程中使用 Web 服务是否是一种正确的方法（以填充列表）？
提前谢谢
编辑= 目前我将“showRecipeDetails”设为静态并在我的自定义 ArrayAdapter 的侦听器中调用它。
convertView.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
RecipeList.showRecipeDetails(position);
}}
我对我的解决方案不满意，我对任何其他解决方案都很感兴趣