Question

这可能是一个简单的问题，我已经看到了类似的问题，但我不知道他们的回答是如何适用于我的，所以我提前道歉但是我收到了这个错误：
不兼容的类型：ArrayList无法转换为int

这是作业，我不确定我做错了什么。

这是我的代码：

package com.example.android.miwok;

import android.os.Bundle;
import android.support.v7.app.AppCompatActivity;
import android.widget.ListView;

import java.util.ArrayList;

public class NumbersActivity extends AppCompatActivity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_numbers);

        // Create a list of words
        ArrayList<Word> words = new ArrayList<Word>();
        words.add(new Word("one", "lutti"));
        words.add(new Word("two", "otiiko"));
        words.add(new Word("three", "tolookosu"));
        words.add(new Word("four", "oyyisa"));
        words.add(new Word("five", "massokka"));
        words.add(new Word("six", "temmokka"));
        words.add(new Word("seven", "kenekaku"));
        words.add(new Word("eight", "kawinta"));
        words.add(new Word("nine", "wo’e"));
        words.add(new Word("ten", "na’aacha"));

        // Create an {@link WordAdapter}, whose data source is a list of {@link Word}s. The
        // adapter knows how to create list items for each item in the list.
        WordAdapter adapter = new WordAdapter(this, words);

        // Find the {@link ListView} object in the view hierarchy of the {@link Activity}.
        // There should be a {@link ListView} with the view ID called list, which is declared in the
        // activity_numbers.xml layout file.
        ListView listView = (ListView) findViewById(R.id.list);

        // Make the {@link ListView} use the {@link WordAdapter} we created above, so that the
        // {@link ListView} will display list items for each {@link Word} in the list.
        listView.setAdapter(adapter);
    }

我收到了这一行的错误：

WordAdapter adapter = new WordAdapter(this, words);

这是我的适配器代码：

package com.example.android.miwok;

        import android.content.Context;
        import android.view.LayoutInflater;
        import android.view.View;
        import android.view.ViewGroup;
        import android.widget.ArrayAdapter;

        import java.util.ArrayList;

/**
 * Created by danle on 7/25/2016.
 */
public class WordAdapter extends ArrayAdapter<Word> {
    public WordAdapter(Context context, int resource) {
        super(context, resource);
    }

    public WordAdapter(NumbersActivity context, int resource, ArrayList<Word> words) {
        super(context, resource, words);
    }




    @Override
    public View getView(int position, View convertView, ViewGroup parent) {
        View listItemView = convertView;
        if (listItemView == null) {
            listItemView = LayoutInflater.from(getContext()).inflate(
                    R.layout.list_item, parent, false);
        }
        return listItemView;
    }
}

Answer 1

问题是你的第二个构造函数。 ArrayAdapter没有超级构造函数只接受上下文和ArrayList<Word>。所以super正在调用(Context, int)并尝试将单词转换为整数资源。

public WordAdapter(NumbersActivity context, ArrayList<Word> words) {
    super(context, words);
}

你的构造函数应该是

public WordAdapter(NumbersActivity context, int resource, ArrayList<Word> words) {
    super(context, resource, words);
}

Answer 2

public WordAdapter(NumbersActivity context, ArrayList<Word> words) {
        super(context, words);
    }

尝试使用对象＆＃34;单词＆＃34;设置ressource-ID。

使用以下构造函数：

ArrayAdapter(Context context, int resource, T[] objects)

See Also

Answer 3

你的构造函数

maj_ax_le

正在调用public WordAdapter(NumbersActivity context, ArrayList<Word> words) { super(context, words); }，它在Super类中没有直接匹配。所以它与super(context, words);相匹配。因此分配和ArrayAdapter(Context context, int resource)到ArrayList问题的错误。您应该使用Super class

中的以下构造函数

int

希望这有帮助

修改

所以你需要用
替换上面提到的代码
ArrayAdapter(Context context, int resource, T[] objects)

并称之为：

public WordAdapter(NumbersActivity context, int resource, ArrayList<Word> words) { super(context, resource, words); }

Answer 4

只需将适配器更改为：

public class WordAdapter extends BaseAdapter {

    ArrayList<Word> items=new ArrayList();

    Context mContext;

    public WordAdapter(Context context,  ArrayList<Word> words) {
        items=words;
        mContext=context;
    }

    @Override
    public int getCount() {
        return items.size();
    }

    @Override
    public Object getItem(int position) {
        return items.get(position);
    }

    @Override
    public long getItemId(int position) {
        return 0;
    }

    @Override
    public View getView(int position, View convertView, ViewGroup parent) {
        Word mWord=items.get(position);
        View listItemView = convertView;
        if (listItemView == null) {
            listItemView = LayoutInflater.from(mContext).inflate(
                    R.layout.list_item, parent, false);
        }
        return listItemView;
    }
}

ArrayList <word>无法转换为int

4 个答案: