Question

我有一个SQLite数据库，其中包含一个名为author的表。我正在尝试提取id和name值并将它们传递给MainActivity，其中id将存储为变量，name将显示在ListView中。

但这就是日志（和我的ListView）的外观：

[com.example.apple.bookshelf.Author@31e44c75,
com.example.apple.bookshelf.Author@24fe640a,
com.example.apple.bookshelf.Author@2c87d47b,
com.example.apple.bookshelf.Author@2ab52298,
com.example.apple.bookshelf.Author@393a3af1,
com.example.apple.bookshelf.Author@2326b6d6]

我可以看到问题是因为我将id和name添加到AuthorList数组，但没有指定要在ListView中显示哪一个。但是我该怎么做？

DatabaseHelper类的片段：

public List<String> getAllAuthors() {

        List authorList = new ArrayList();
    // Select all query
    String selectQuery = "SELECT * FROM " + AUTHORS + " ORDER BY name_alphabetic";

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);

    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {
            Author author = new Author();
            author.setID(Integer.parseInt(cursor.getString(0)));
            author.setName(cursor.getString(1));
            authorList.add(author);
        } while (cursor.moveToNext());
    }

    // return author list
    return authorList;
}

作者类：

public class Author {

    int id;
    String name;
    String name_alphabetic;

    public Author() {

    }

    public Author(int id, String name, String name_alphabetic) {
        this.id = id;
        this.name = name;
        this.name_alphabetic = name_alphabetic;
    }

    // getters

    public int getID() {
        return this.id;
    }

    public String getName() {
        return this.name;
    }

    public String getNameAlphabetic() {
        return this.name_alphabetic;
    }

    // setters

    public void setID(int id) {
        this.id = id;
    }

    public void setName(String name) {
        this.name = name;
    }

    public void setNameAlphabetic(String name_alphabetic) {
        this.name_alphabetic = name_alphabetic;
    }
}

来自MainActivity的

片段：

        // connect authorsListView variable to XML ListView
        authorsListView = (ListView) findViewById(R.id.authors_list_view);

        // create new database helper
        DatabaseHelper db = new DatabaseHelper(this);

        // create new list through getAllAuthors method (in DatabaseHelper class)
        List authorList = db.getAllAuthors();

        Log.i("authors", authorList.toString());

        // create new Array Adapter
        ArrayAdapter<String> arrayAdapter =
                new ArrayAdapter<String>(this, android.R.layout.simple_list_item_1, authorList);

        // link ListView and Array Adapter
        authorsListView.setAdapter(arrayAdapter);

修改

我使用下面评论中的this thread解决了问题。将此代码添加到我的Author类以覆盖从toString()类继承的Object方法可以解决问题：

@Override
  public String toString() {
    return name;
  }

Answer 1

您并没有真正使用Author课程。我建议您执行以下操作：

将您的List更改为List<Author>：

public List<Author> getAllAuthors() {

    List<Author> authorList = new ArrayList<>();
    // Select all query
    String selectQuery = "SELECT * FROM " + AUTHORS + " ORDER BY name_alphabetic";

    SQLiteDatabase db = this.getWritableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);

    // looping through all rows and adding to list
    if (cursor.moveToFirst()) {
        do {
            Author author = new Author();
            author.setID(Integer.parseInt(cursor.getString(0)));
            author.setName(cursor.getString(1));
            authorList.add(author);
        } while (cursor.moveToNext());
    }

    // return author list
    return authorList;
}

然后在MainActivity.java中更改ArrayAdapter：

的行

ArrayAdapter<Author> arrayAdapter =
            new ArrayAdapter<>(this, android.R.layout.simple_list_item_1, authorList);

确保toString() Author返回按您希望的方式格式化的字符串。

在游标中设置多个变量不符合预期

1 个答案: