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即使用户只输入首字母，我如何过滤结果。例如，如果列表中的项目是“George Washington”，如果用户在搜索框中输入“g w”，它会显示结果。下面是我的课

public void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);

    // Listview Data
    String products[] = {"Dell Inspiron", "HTC One X", "HTC Wildfire S", "HTC Sense", "HTC Sensation XE",
                            "iPhone 4S", "Samsung Galaxy Note 800",
                            "Samsung Galaxy S3", "MacBook Air", "Mac Mini", "MacBook Pro"};

    lv = (ListView) findViewById(R.id.list_view);
    inputSearch = (EditText) findViewById(R.id.inputSearch);

    // Adding items to listview
    adapter = new ArrayAdapter<String>(this, R.layout.list_item, R.id.product_name, products);
    lv.setAdapter(adapter);

    /**
     * Enabling Search Filter
     * */
    inputSearch.addTextChangedListener(new TextWatcher() {

        @Override
        public void onTextChanged(CharSequence cs, int arg1, int arg2, int arg3) {
            // When user changed the Text
            MainActivity.this.adapter.getFilter().filter(cs);   
        }

        @Override
        public void beforeTextChanged(CharSequence arg0, int arg1, int arg2,
                int arg3) {
            // TODO Auto-generated method stub

        }

        @Override
        public void afterTextChanged(Editable arg0) {
            // TODO Auto-generated method stub                          
        }
    });
   }    
   }
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1 回答 1

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您需要设置一个Content Provider并在每次用户输入字母时向该提供者查询结果。

于 2013-08-15T17:22:38.780 回答