Question

我已经仔细研究了许多类似的问题，但我无法找到我想要的答案。我几乎是自学成才，所以我为这个问题道歉。基本上我试图访问我的数组中的字符串ID并将其转换为可读的字符串，以便我可以将其发送到谷歌地图，以便用户可以在那里导航。这是我得到的最接近的。

import android.content.Intent;
import android.net.Uri;
import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.AdapterView;
import android.widget.ListView;

import java.util.ArrayList;

public class FoodFragment extends Fragment {

    public FoodFragment() {

    }


    @Override
    public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
        View rootView = inflater.inflate(R.layout.word_list, container, false);

        final ArrayList<Location> places = new ArrayList<>();
        places.add(new Location(R.string.food1, R.string.desc1, R.drawable.tapass));
        places.add(new Location(R.string.food2, R.string.desc2, R.drawable.baguette));
        places.add(new Location(R.string.food3, R.string.desc3, R.drawable.sushi));
        places.add(new Location(R.string.food4, R.string.desc4, R.drawable.burger));
        places.add(new Location(R.string.food5, R.string.desc5, R.drawable.meat));
        places.add(new Location(R.string.food6, R.string.desc6, R.drawable.pizza));

        final Location here = places.get(0);
        final int address = here.getName();

        MyAdapter adapter = new MyAdapter(getActivity(), places);

        final ListView listView = (ListView) rootView.findViewById(R.id.list);

        listView.setAdapter(adapter);

        listView.setOnItemClickListener(new AdapterView.OnItemClickListener() {
            @Override
            public void onItemClick(AdapterView<?> adapterView, View view, int position, long l) {

                Intent intent = new Intent(android.content.Intent.ACTION_VIEW,
                        Uri.parse("google.navigation:q=" + address));
                startActivity(intent);
            }
        });
        return rootView;
    }
}

getName来自我的Location类。我意识到它返回一个int（这样我可以将string.xml中的字符串id放入数组中）但是如何让它返回实际的字符串值呢？

public class Location {

    private int mName;
    private int mDescription;
    private int mImage;


    public Location (int name, int description, int image){
        mName = name;
        mDescription = description;
        mImage = image;
    }

    public int getName(){
        return mName;
    }

    public int getDescription(){
        return mDescription;
    }

    public int getImage(){
        return mImage;
    }
}

我再次为noob问题道歉，非常感谢任何帮助。

Answer 1

您应该将Location类更改为此类

public class Location {

    private int mName;
    private int mDescription;
    private int mImage;
    private Context mContext;


    public Location (int name, int description, int image, Context context){
        mName = name;
        mDescription = description;
        mImage = image;
        mContext = context;
    }

    public String getName() {
        return mContext.getString(mName)
    }

    public int getDescription(){
        return mDescription;
    }

    public int getImage(){
        return mImage;
    }
}

然后您可以像这样创建位置实例

places.add(new Location(R.string.food1, R.string.desc1, R.drawable.tapass, getActivity()));

Answer 2

此处address是id的资源string.xml 您可以使用

访问address id的值

 String value = getString(address);

如何将ArrayList中的int String id转换为可用的字符串？

2 个答案: