将ArrayList转换为2D数组

In Java how do you convert a ArrayList into a two dimensional array Object[][]?

在Java中如何将ArrayList转换为二维数组Object [] []?

From comments: I will describe you the problem with more details: an XML file includes a list of contacts (e.g. name, address...). The only way I can obtain this information is through an ArrayList, which will be given to me. As I need to store the content of this array list in a Java Swing table in an ordered manner, I was thinking to convert it into a two dimensional array of objects

来自评论:我将通过更多详细信息向您描述问题:XML文件包含联系人列表(例如姓名,地址......)。我可以获取此信息的唯一方法是通过ArrayList,它将提供给我。因为我需要以有序的方式将这个数组列表的内容存储在Java Swing表中,所以我想将它转换为二维对象数组

7 个解决方案

#1

I presume you are using the JTable(Object[][], Object[]) constructor.

我假设您正在使用JTable(Object [] [],Object [])构造函数。

Instead of converting an ArrayList<Contact> into an Object[][], try using the JTable(TableModel) constructor. You can write a custom class that implements the TableModel interface. Sun has already provided the AbstractTableModel class for you to extend to make your life a little easier.

不要将ArrayList 转换为Object [] [],而是尝试使用JTable(TableModel)构造函数。您可以编写实现TableModel接口的自定义类。 Sun已经提供了AbstractTableModel类供您扩展,以使您的生活更轻松。

public class ContactTableModel extends AbstractTableModel {

    private List<Contact> contacts;

    public ContactTableModel(List<Contact> contacts) {
        this.contacts = contacts;
    }

    public int getColumnCount() {
        // return however many columns you want
    }

    public int getRowCount() {
        return contacts.size();
    }

    public String getColumnName(int columnIndex) {
        switch (columnIndex) {
        case 0: return "Name";
        case 1: return "Age";
        case 2: return "Telephone";
        // ...
        }
    }

    public Object getValueAt(int rowIndex, int columnIndex) {
        Contact contact = contacts.get(rowIndex);

        switch (columnIndex) {
        case 0: return contact.getName();
        case 1: return contact.getAge();
        case 2: return contact.getTelephone();
        // ...
        }
    }

}

Later on...

List<Contact> contacts = ...;
TableModel tableModel = new ContactTableModel(contacts);
JTable table = new JTable(tableModel);

#2

The simple way is to add a method to the Contact like this:

简单的方法是向Contact添加一个方法,如下所示:

public Object[] toObjectArray() {
    return new Object[] { getName(), getAddress, /* ... */ };
}

and use it like this:

并像这样使用它:

ArrayList<Contact> contacts = /* ... */
Object[][] table = new Object[contacts.size()][];
for (int i = 0; i < contacts.size(); i++) {
    table[i] = contacts.get(i).toObjectArray();
}

#3

Try this:

ArrayList<Integer> list = new ArrayList<Integer>();
list.add(..);
list.add(..);
list.add(..);
list.add(..);
list.add(..);
list.add(..);
int[][] a = new int[list.size()][list.size()];
    for(int i =0; i < list.size(); i++){
      for(int j =0; j <list.size(); j++){
        a[i][j]= list.get(j +( list.size() * i));
      }
  }

#4

I managed to find "a way" to do so, knowing the number of attributes each contacts has (6). So considering an ArrayList listofContacts

我知道每个联系人拥有的属性数量,我设法找到“一种方法”(6)。所以考虑一个ArrayList listofContacts

    int numberOfContacts = listofContacts.size()/6;
    Object[][] newArrayContent = new Object[numberOfContacts][6];

    for(int x = 0; x<numberOfContacts; x++){
        for(int z = 0; z < 6; z++){
        int y = 6 * x;
        newArrayContent [x][z] = list.get(y+z); 
        System.out.println(newArrayContent [x][z].toString());
        }
    }

#5

What you really want is to sort the ArrayList. To do that your Contacts class must implement a Comparator method.

你真正想要的是对ArrayList进行排序。为此,您的Contacts类必须实现Comparator方法。

Check the next page for an example: http://www.java-examples.com/sort-java-arraylist-descending-order-using-comparator-example

查看下一页的示例:http://www.java-examples.com/sort-java-arraylist-descending-order-using-comparator-example

#6

I will recommend that you parse your XML into java objects and store the object in a custom data object. This will make it easier for you to do many operations on the available data.

我建议您将XML解析为java对象并将对象存储在自定义数据对象中。这将使您更容易对可用数据执行许多操作。

Here is small tutorial on how to do it.

这是关于如何做到这一点的小教程。

#7

public static String[][] convertListIntoArrayObj(List<TeamMenuSelected> possibilities) {
    int numberOfColums = 2;
    int numberOfRows = possibilities.size();
    String[][] values = new String[numberOfRows][numberOfColums];

    for(int x=0; x<possibilities.size(); x++) {
        TeamMenuSelected item = possibilities.get(x);
        values[x][0] = item.getTeamName();
        values[x][1] = item.getTeamCuisine();
    }

    return values;
}

#1