根据名称在Android中过滤Listview [英] Filtering Listview according to name in android

问题描述

我正在android上做一个小工作.它有一个列表视图，其中包含一些朋友的名字和他们的生日.有一个搜索文本框来过滤列表视图.我还添加了过滤功能.一个名称可能包含3个部分(第一，中间，最后)，所以我想根据名称的每个部分的首字母过滤列表视图.我的代码现在仅在全名包含键入字母时过滤.考虑一个示例，我有2个朋友，名称是Shezan Mahmud，Bob Ericson.如果我在搜索文本框中输入字母"e"，我的代码将显示包含这两个名称的listview输出(因为两个名称都包含字母"e").但是我想要包含名称Bob Ericson的输出(因为该名称的第二部分包含字母"e"作为第一个字母).任何建议如何完成?我的代码在这里..

I am doing a small work on android.It has a listview which contains some friend''s name and their birthday.There is a search textbox to filter the listview.I also add filtering functionality.As a name may contain 3 parts(first,middle,last),so i want to filter the listview according to first letter of every part of name.My code now filters only when the whole name contains the typed letter.Consider an example,I have 2 friends which name is Shezan Mahmud,Bob Ericson.if i type a letter ''e'' in the search textbox my code shows the listview output containing those 2 names(as both 2 names contain the letter ''e''). But i want the output which contains the name,Bob Ericson(because 2nd part of this name contains the letter ''e'' as first letter).Any suggestion how can it be done?My code''s here..

	// The view in this tabbed example
	private ListView ls1;
	 
	private TabHost tabHost;
        public static String[] friends={"Shuvro Pal","Tamanna Hossain Trena","Ariful Islam","Nargis Rahman"};
	        public static String[] birthdate={"1989","1989","1990","1985"};

	        private EditText filterText;
	        private ArrayList<String> arr_sort= new ArrayList<String>();
	        int textLength=0;

    /** Called when the activity is first created. */
	        @Override
	public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.main);
 

        filterText = (EditText) findViewById(R.id.search_box);

        	        tabHost = getTabHost();
        	        tabHost.setOnTabChangedListener(this);
        	 
        	        // setup list view 1
        	        ls1 = (ListView) findViewById(R.id.list1);
       
        	        final List<Model> list = new ArrayList<Model>();

        	        final ArrayAdapter<Model> adapter = new MyCustomArrayAdapter(this, list);
        	        ls1.setAdapter(adapter);
        	        
        	        for(int i=0;i<friends.length;i++)
        	        list.add(get(friends[i],birthdate[i]));
        	        
     filterText.addTextChangedListener(new TextWatcher() 
     {
	@Override
	public void onTextChanged(CharSequence s, int start, int before, int count) {
// TODO Auto-generated method stub
textLength=filterText.getText().length();
list.clear();
for(int i=0;i<friends.length;i++)
{
  if(textLength<=friends[i].length())
  {
	String[] word=friends[i].split(" ");
	for(int j=0;j<word.length;j++)
	{
               if(friends[i].toLowerCase().contains(filterText.getText().toString().toLowerCase()))
{
											list.add(get(friends[i],birthdate[i]));
	break;
	}
      }
    }
   }
     ls1.setAdapter(adapter);
 }

@Override
public void beforeTextChanged(CharSequence s, int start, int count,int after)
{
public void afterTextChanged(Editable s) 
{
}
});
        	        
 // add views to tab host
        tabHost.addTab(tabHost.newTabSpec(LIST1_TAB_TAG).setIndicator(LIST1_TAB_TAG).setContent(new TabContentFactory() 
        {
            public View createTabContent(String arg0) 
            {
	                return ls1;
            }
         }));

推荐答案

我自己完成的.用空格分隔名称，然后检查键入的字符串的索引. >

I have done it myself.Splitting a name by space and then checking the index of the typed string..

String[] word1=friends[i].split(" ");

for(int a=0;a<word1.length;a++)
{
   index=word1[a].toLowerCase().indexOf(filterText.getText().toString().toLowerCase());

    if(index == 0)
    {
      list.add(get(friends[i],birthdate[i]));
      break;
    }

这篇关于根据名称在Android中过滤Listview的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！