Java ListView正在填充相同的值

发生的情况是，我有一个ArrayList，当我循环该列表以使用列表适配器将该对添加到具有两个独立列的listview时，我尝试将该对的第一个字符串添加到某个textview，然后将第二个字符串添加到某个textview，然后它在整个列表中循环，然后向所有行添加相同的值

列表适配器的代码：

package com.example.test2;

import android.content.Context;
import android.util.Log;
import android.util.Pair;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.ArrayAdapter;
import android.widget.TextView;
import java.util.ArrayList;


public class ThreeColumn_ListAdapter extends ArrayAdapter<Pair<String, String>> {
    private LayoutInflater mInflater;
    private ArrayList<Pair<String, String>> news;
    private int mViewResourceId;


    public ThreeColumn_ListAdapter(Context context, int textViewResourceId, ArrayList<Pair<String, String>> news) {
        super(context, textViewResourceId, news);
        this.news = news;
        mInflater = (LayoutInflater) context.getSystemService(Context.LAYOUT_INFLATER_SERVICE);
        mViewResourceId = textViewResourceId;
    }

    public View getView(int position, View convertView, ViewGroup parent) {
        convertView = mInflater.inflate(mViewResourceId, null);
        TextView firstName = convertView.findViewById(R.id.textFirstName);
        TextView lastName = convertView.findViewById(R.id.textLastName);
        //TextView favFood = convertView.findViewById(R.id.textFavFood);

        Log.i("n2",news.toString());
        for(int j=0;j<news.size();j++){
           firstName.setText(news.get(j).first);
           lastName.setText(news.get(j).second);

        }


        return convertView;
    }

    }

在listview布局中查看的输出：

---

Pair{jpy 1.8649}

在textview

firstName

中Pair的第一个值迭代了8次，在textview

lastName

中Pair的第二个值迭代了8次，8是

ArrayList news

的长度，我所要做的就是删除for循环，只做news.get（position）。首先，不要对新闻执行循环，因为listview自然执行循环

因此，以前：

public View getView(int position, View convertView, ViewGroup parent) {
        convertView = mInflater.inflate(mViewResourceId, null);
        TextView firstName = convertView.findViewById(R.id.textFirstName);
        TextView lastName = convertView.findViewById(R.id.textLastName);
        //TextView favFood = convertView.findViewById(R.id.textFavFood);

        Log.i("n2",news.toString());
        for(int j=0;j<news.size();j++){
           firstName.setText(news.get(j).first);
           lastName.setText(news.get(j).second);

        }


        return convertView;
    }

---

Log.Ith中的“n2”是存储在新闻列表中的内容，因此第二个代码显示列表；[对价{美元-0.8068}，对价{欧元0.5327}，对价{1.2172}，对价{nzd-2.7538}，对价{cad 0.7586}，对价{aud-1.7719}，对价{chf 0.9591}，对价{1.8649}]

public View getView(int position, View convertView, ViewGroup parent) {
        convertView = mInflater.inflate(mViewResourceId, null);
        TextView firstName = convertView.findViewById(R.id.textFirstName);
        TextView lastName = convertView.findViewById(R.id.textLastName);
        //TextView favFood = convertView.findViewById(R.id.textFavFood);

        Log.i("n2",news.toString());
        for(int j=0;j<news.size();j++){
           firstName.setText(news.get(j).first);
           lastName.setText(news.get(j).second);

        }


        return convertView;
    }

 public View getView(int position, View convertView, ViewGroup parent) {
        convertView = mInflater.inflate(mViewResourceId, null);
        TextView firstName = convertView.findViewById(R.id.textFirstName);
        TextView lastName = convertView.findViewById(R.id.textLastName);
        //TextView favFood = convertView.findViewById(R.id.textFavFood);
        firstName.setText(news.get(position).first);
        lastName.setText(news.get(position).second);


        return convertView;
    }