Android 在 ViewPager 上方显示 SearchView 结果 ListView

Question

您好，我想通过SearchView 小部件在我的应用中搜索用户。当SearchWidget 处于活动状态时，我想在相同XML 布局的ViewPager 上方的ListView 中显示建议的用户名。 我使用了以下 xml

<?xml version="1.0" encoding="utf-8"?><RelativeLayout xmlns:android="http://schemas.android.com/apk/res/android"
android:layout_width="fill_parent"
android:layout_height="fill_parent" >    
<ListView
    android:id="@+id/searched_friends_list"
    android:layout_width="fill_parent"
    android:layout_height="wrap_content"
    android:divider="@color/list_divider"
    android:dividerHeight="1dp"
    android:layout_marginLeft="15dp"
    android:layout_marginRight="15dp"
    android:visibility="gone"
    android:listSelector="@drawable/list_row_selector" />       
<android.support.v4.view.ViewPager
    android:id="@+id/pager"
    android:layout_width="match_parent"
    android:layout_height="match_parent" >
</android.support.v4.view.ViewPager>

在代码中我可见ListView 以向用户展示。

@Override
public boolean onQueryTextChange(String newText) {
    friendListView.setVisibility(View.VISIBLE);
    if (TextUtils.isEmpty(newText)) {
        friendListView.clearTextFilter();
    } else {
        friendListView.setFilterText(newText.toString());
    }
    return true;
}

它不显示ListView

但是当我添加以下代码时

@Override
public boolean onQueryTextChange(String newText) {
    friendListView.setVisibility(View.VISIBLE);
    viewPager.setVisibility(View.GONE);
    if (TextUtils.isEmpty(newText)) {
        friendListView.clearTextFilter();
    } else {
        friendListView.setFilterText(newText.toString());
    }
    return true;
}

它显示以下输出

但我想要ListView 高于ViewPager。提前致谢。

Answer 1

如果您希望您的 ListView 实时显示用户将在 searchView 中建议的结果 - 请在您的适配器中使用过滤器。

添加到适配器：

ArrayList<full_data> adapterDataList = new ArrayList<full_data>();
ArrayList<filtered_data> mFilteredDataList = new ArrayList<filtered_data>(listOR);
Filter mFilter = new Filter() {
@Override
protected FilterResults performFiltering(CharSequence searchTerm) {
Locale locale = Locale.getDefault();
FilterResults results = new FilterResults();
if ((searchTerm null) || (searchTerm.length() 0)) {
ArrayList<String> searchList = new ArrayList<String>(adapterDataList);
results.values = searchList;
results.count = searchList.size();
} else {
final String searchTermString = searchTerm.toString().toLowerCase(locale);
final String[] words = searchTermString.split(" ");
final int wordCount = words.length;
final ArrayList<String> newValues = new ArrayList<String>();

for (final String value : adapterDataList) {
                    if (value == null) {
                        continue;
                    }
                    final String valueText = value.toLowerCase(locale);
for (int k = 0; k < wordCount; k++) {
                        if (valueText.contains(words[k])) {
                            newValues.add(value);
                            break;
                        }
                    }
                }
                results.values = newValues;
                results.count = newValues.size();
            }
            return results;
        }
@Override
        protected void publishResults(CharSequence constraint, FilterResults results) {
            mFilteredDataList = (ArrayList&lt;String&gt;)results.values;
            notifyDataSetChanged();
        }
    };
public Filter getFilter(){ return mFilter; }

to SearchView's TextChangedListener:
@Override
public void afterTextChanged(Editable searchText) {
mAdapter.getFilter().filter(searchText);
}

我认为它应该可以正常工作。