Question

我希望有一个搜索字段，我可以在其中放置我想要搜索的内容。我已经有了这个实现。

import java.util.ArrayList;
import java.util.HashMap;

import android.app.Activity;
import android.os.Bundle;
import android.text.Editable;
import android.text.TextWatcher;
import android.widget.ArrayAdapter;
import android.widget.EditText;
import android.widget.ListView;

public class MainActivity extends Activity {

    // List view
    private ListView lv;

    // Listview Adapter
    ArrayAdapter<String> adapter;

    // Search EditText
    EditText inputSearch;


    // ArrayList for Listview
    ArrayList<HashMap<String, String>> productList;

    @Override
    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                          
            }
        });
    }    
}

但是如果搜索字段以相同的char序列开头，它只返回结果。例如，如果我写“Galaxy Note”，它不会返回任何结果，即使有一个名为“Samsung Galaxy Note 800”的元素

Answer 1

您应该编写自己的过滤器类，并在baseAdapter的getFilter（）函数中使用它。

private class yourFilter extends Filter
{

    @Override
    protected FilterResults performFiltering(CharSequence constraint) {
        constraint = constraint.toString().toLowerCase();
        FilterResults result = new FilterResults();
        ArrayList<String> f = new ArrayList<String>();
        ArrayList<String> p = new ArrayList<String>();
        for(String product : products){
               p.add(product);
            }
        if(constraint != null && constraint.toString().length() > 0)
        {
            for(int i = 0; i < p.size(); i++)
            {
                String product = p.get(i);
                if(product.toLowerCase().contains(constraint))
                    f.add(product);
            }
            result.count = f.size();
            result.values = f;
        }
        else
        {              
             result.values = p;
             result.count = p.size();                
        }
        return result;
    }


    @Override
    protected void publishResults(CharSequence constraint, FilterResults results) {
        yourFilteredData = (ArrayList<String>)results.values;
        notifyDataSetChanged(); 
    }

}

Answer 2

看到您的ListView包含纯字符串值，那么实现此目的的最简单方法是使用：

lv.setTextFilterEnabled(true);

初始化lv变量后。但是，如果ListView包含复杂（非原始）数据类型，则此方法将无法按预期工作。请参阅Google's documentation以获取帮助。

在Android中按字符串过滤搜索结果

2 个答案: