如果你试图返回一个新的对象，你必须告诉它的属性，

select new {PlayerOne = a.PlayerOne
            //More Properties Here
           }

根据你的例子，

var db = new DataClasses1DataContext();
    var history =( from a in db.ResultsHistories
        where
            a.PlayerOne == _playerOne && a.PlayerTwo == comboBox.Text ||
            a.PlayerTwo == _playerOne && a.PlayerOne == comboBox.Text
        select new
        {
            PlayerOne = a.PlayerOne,
            PlayerTwo = a.PlayerTwo,
            Date = a.Date,
            ResultOne = a.ResultOne,
            ResultTwo = a.ResultTwo
        }).ToList();

    listBox.ItemsSource = history;

这将返回一个没有类型的新对象列表，但属性将在那里

编辑：如果您想要表中的所有数据，其中PlayerOne和PlayerTwo等于一个文本框值，请执行以下操作

var history = db.ResultsHistories.Where(x => x.PlayerTwo == comboBox.Text || x.PlayerOne == comboBox.Text).Select(x => x).ToList();

尝试添加此内容

• “，“+语句之间

a.PlayerOne+“、“+a.PlayerTwo+“、“+ a.Date+“、“+ a.ResultOne+“、“+ a.ResultTwo

它会工作

<ListBox x:Name="listBx_ShowPersons">
                <ListBox.ItemTemplate>
                    <DataTemplate>
                        <StackPanel Orientation="Horizontal">
                            <TextBlock Text="{Binding firstName}"/>
                            <TextBlock Text=", "/>
                            <TextBlock Text="{Binding lastName}"/>
                        </StackPanel>
                    </DataTemplate>
                </ListBox.ItemTemplate>                    
            </ListBox>

listBx_ShowPersons.ItemsSource = from p in _dataContext.Persons select new { p.firstName, p.lastName };

请在这里看看答案：Multicolumn ListBox in WPF
您也可以使用DataGrid来实现此目的。XAML代码是：

<DataGrid x:Name="NamesDataGrid"></DataGrid>

后面有一个非常简单的代码：

public partial class MainWindow : Window
{
    private readonly List<Name> _nameList = new List<Name>
    {
        new Name("Eric","Clapton"),
        new Name("Mark", "Knopfler")
    };
    public MainWindow()
    {
        InitializeComponent();
        NamesDataGrid.ItemsSource = _nameList;
    }
}

internal class Name
{
    public Name(string firstName, string lastName)
    {
        FirstName = firstName;
        LastName = lastName;
    }
    public string FirstName { get; set; }
    public string LastName { get; set; }
}

它看起来像这样：