我需要在 ListView 中搜索项目。我有简单的字符串数组和数组适配器,放置我在一些不同问题及其工作中找到的代码!但我有一个问题,这个搜索非常慢并且滞后一个字母.我有单词“arbuz”(一个代表许多其他单词),我打印了一个-什么也没有,另一个字母-和单词搜索“a”,另一个-并且没有这样的单词。我希望你理解我。我如何改进这个搜索? 我的代码:
public class FragmentWithList extends ListFragment {
ListView mainList;
EditText inputSearch;
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
View v = inflater.inflate(R.layout.fragment_with_list, null);
mainList = (ListView) v.findViewById(android.R.id.list);
inputSearch = (EditText)v.findViewById(R.id.inputSearch);
return v;
}
public void onCreate(Bundle savedInstanceState){
setHasOptionsMenu(true);
super.onCreate(savedInstanceState);
}
public void onActivityCreated(Bundle savedInstanceState) {
super.onActivityCreated(savedInstanceState);
String values[] = new String[]{"arbuz","abrikos","banan","cup","charlie","derevo","grusha","ogurets","sliva","jabloko","volk","dom","igra","pharaon","muscle"};
Comparator<String> ALPHABETICAL_ORDER1 = new Comparator<String>() {
public int compare(String object1, String object2) {
int res = String.CASE_INSENSITIVE_ORDER.compare(object1.toString(), object2.toString());
return res;
}
};
Collections.sort(Arrays.asList(values), ALPHABETICAL_ORDER1);
final ArrayAdapter<String> adapter = new ArrayAdapter<String>(getActivity(),android.R.layout.simple_list_item_1,values);
mainList.setAdapter(adapter);
inputSearch.addTextChangedListener(new TextWatcher() {
@Override
public void beforeTextChanged(CharSequence s, int start, int count, int after) {
adapter.getFilter().filter(s.toString());
}
@Override
public void onTextChanged(CharSequence s, int start, int before, int count) {
}
@Override
public void afterTextChanged(Editable s) {
}
});
}
}
最佳答案
在afterTextChanged()
方法中添加以下代码
adapter.getFilter().filter(s.toString());
关于java - 使用 edittext 在 ListView 中搜索,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/34658920/