Question

是否可以使用SelectMany（）并使其行为类似于左连接？

我正在尝试将实体对象压缩为表格格式。只要有一个子对象，SelectMany（）的工作方式就像一个魅力，但我希望看到所有的父对象，无论它是否有子节点。

public class Folder
{
    [Key]
    public int Folder_Id { get; set; }
    public string Folder_Name { get; set; }
    public int Folder_order { get; set; }
    public virtual ICollection<Module> Modules { get; set; }
}
public class Module
{        
    [Key]
    public int Module_Id { get; set; }
    public int? Folder_Id { get; set; }
    public string ModuleName { get; set; }
    public virtual Folder Folder { get; set; }

}
public static List<Menu> GetMenu()
{
    var Menu = db.Folder.Get();
    var q = from u in Menu
    from m in u.Modules
    select new {folderName = u.Folder_Name, moduleName = m.ModuleName};
    return q;
}

我需要的是：

FOLDERNAME =＆＃34; ENG＆＃34;，MODULENAME =＆＃34; REQ＆＃34;

FOLDERNAME =＆＃34; ENG＆＃34;，MODULENAME =＆＃34; BOM＆＃34;

FOLDERNAME =＆＃34;收件箱＆＃34;，MODULENAME =＆＃34;＆＃34;

FOLDERNAME =＆＃34;帮助＆＃34;，MODULENAME =＆＃34;＆＃34;

FOLDERNAME =＆＃34;信息中心＆＃34;，MODULENAME =＆＃34;＆＃34;

FOLDERNAME =＆＃34;采购＆＃34;，MODULENAME =＆＃34;缩进＆＃34;

Answer 1

当你在连接上选择导航属性时，你确实遵循了实体框架的推荐做法。唉，导航属性仅适用于内连接。因此，在这种情况下，SelectMany对您无济于事。所以你必须求助于加入：

var q =
    from u in Menu
    join m in Modules on u.Folder_Id equals m.Folder_Id into modules
    from mm in modules.DefaultIfEmpty()
    select new {folderName = u.Folder_Name, moduleName = mm?.ModuleName ?? String.Empty};
    return q;

使用SelectMany（）的平面数据，如左连接

1 个答案: