我有以下型号:
public class ApplicationUser : IdentityUser<string, ApplicationUserLogin, ApplicationUserRole, ApplicationUserClaim>
{
public bool HasPermission(string permission)
{
return Roles.Any(r => r.Role.Permissions
.Any(p => p.Name == permission));
}
// Some other Stuff
}
public class ApplicationRole : IdentityRole<string, ApplicationUserRole>
{
public ApplicationRole(string name) : this()
{
this.Name = name;
}
public virtual ICollection<Permission> Permissions { get; set; }
}
public class ApplicationUserRole : IdentityUserRole
{
public ApplicationUserRole(): base() { }
public virtual ApplicationRole Role { get; set; }
}
public class Permission
{
public byte Id { get; set; }
public string Name { get; set; }
public virtual ICollection<ApplicationRole> Roles { get; set; }
}
基本上它们映射到五个表:Users,Roles,UserRoles,Permissions和RolePermissions - 目前我没有RolePermission的模型 - 我需要一个吗?
我希望使用带有lambda表达式的linq获取给定用户的权限列表,类似于以下SQL:
Select p.Id, p.Name
From Permission p
inner join RolePermission rp on rp.PermissionId = p.Id
inner join UserRole ur on ur.RoleId = rp.RoleId
where ur.UserId = 'xyz'
我不确定我从哪个实体开始使用LINQ获取权限列表 - 我只能设法使用以下内容获取角色?
var permissions = _context.Users.Where(u => u.Id == userId)
.SelectMany(r => r.Roles.SelectMany(p => p.Role.Permissions))
.ToList();
有什么想法吗?
答案 0 :(得分:0)
要在Linq中加入,你可以使用类似的东西。
var permissions = new List<Permission>(){
new Permission (){
Id= 1,
Name = "Test1"
},
new Permission{ Id = 2,
Name = "Test2"}
};
var applicationRoles = new List<ApplicationRole>(){
new ApplicationRole{
Id =1,
PermissionId = 2
}};
var x = from p in permissions
join ar in applicationRoles on p.Id equals ar.PermissionId
select p;
public class Permission{
public byte Id {get;set;}
public string Name {get;set;}
public List<ApplicationRole> Roles {get;set;}
}
public class ApplicationRole {
public byte Id {get;set;}
public byte PermissionId {get;set;}
}
我不确定这是不是你真正要问的,所以我希望它可以帮到你。
答案 1 :(得分:0)
管理如下:
var permissions = _context.Users.Where(u => u.Id == userId)
.SelectMany(r => r.Roles.SelectMany(p => p.Role.Permissions)).Distinct().ToList();