I want the MachineRecord to contain a GroupRecord Id, like the following:
MachineRecord
Id
Name
Number
GroupRecord
GroupRecord
Id
Name
What I have tried:
I have tried to make the property in the model a
public virtual GroupRecord GroupRecord { get; set; }
and the database column name GroupRecord_Id, but somehow that won’t work, tried some other ways like giving the property the name GroupRecordId and the table column GroupRecord and so on, but all with no results.
So the question is, how to get this to work so the MachineRecord table contains a GroupRecord Id?
Machine.cs
namespace PowerAll.Voorraad.Models
{
public class MachineRecord
{
public virtual int Id { get; set; }
public virtual int MachineNumber { get; set; }
public virtual string Title { get; set; }
public virtual string Description1 { get; set; }
public virtual string Description2 { get; set; }
public virtual string Description3 { get; set; }
public virtual string Description4 { get; set; }
public virtual string Description5 { get; set; }
public virtual string Description6 { get; set; }
public virtual string SerialNumber { get; set; }
public virtual GroupRecord GroupRecord { get; set; }
public virtual char PriceType { get; set; }
public virtual decimal Price { get; set; }
public virtual int Year { get; set; }
}
}
Group.cs
namespace PowerAll.Voorraad.Models
{
public class GroupRecord
{
public virtual int Id { get; set; }
public virtual string GroupName { get; set; }
}
}
Migrations.cs
namespace PowerAll.Voorraad
{
public class Migrations : DataMigrationImpl
{
public int Create()
{
SchemaBuilder.CreateTable("GroupRecord", table => table
.Column<int>("Id", column => column.PrimaryKey().Identity())
.Column<string>("GroupName", column => column.WithLength(100))
);
SchemaBuilder.CreateTable("MachineRecord", table => table
.Column<int>("Id", column => column.PrimaryKey().Identity())
.Column<int>("MachineNumber", column => column.NotNull())
.Column<string>("Title", column => column.NotNull().WithLength(40))
.Column<string>("Description1", column => column.WithLength(70))
.Column<string>("Description2", column => column.WithLength(70))
.Column<string>("Description3", column => column.WithLength(70))
.Column<string>("Description4", column => column.WithLength(70))
.Column<string>("Description5", column => column.WithLength(70))
.Column<string>("Description6", column => column.WithLength(70))
.Column<string>("SerialNumber", column => column.WithLength(20))
.Column<int>("GroupRecord_id", column => column.NotNull())
.Column<char>("PriceType", column => column.NotNull().WithLength(1))
.Column<decimal>("Price", column => column.NotNull())
.Column<int>("Year", column => column.WithLength(4))
);
// Return the version that this feature will be after this method completes
return 1;
}
}
}
MachineController.cs
namespace PowerAll.Voorraad.Controllers
{
[Themed]
public class MachineController : Controller
{
private readonly IRepository<MachineRecord> machineRecords;
public MachineController(IRepository<MachineRecord> MachineRecords) {
machineRecords = MachineRecords;
}
public ActionResult List()
{
var items = machineRecords.Table;
return View(items);
}
}
}
EDIT 2:
I came this far thanks to endorphin, but when I try to execute AddDummyData() I get this error: could not insert: [PowerAll.Voorraad.Models.MachineRecord][SQL: INSERT INTO PowerAll_Voorraad_MachineRecord (MachineNumber, Title, Description1, Description2, Description3, Description4, Description5, Description6, SerialNumber, PriceType, Price, Year) VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?); select SCOPE_IDENTITY()] While when I step trough the code all values are being posted right.
This is my AddDummyData():
public void AddDummyData()
{
var GroupRecord = new GroupRecord { GroupName = "Oldtimers" };
groupRecords.Create(GroupRecord);
var MachineRecord = new MachineRecord
{
MachineNumber = 100001,
Title = "Landini L25",
Description1 = "Desc 1",
Description2 = "Desc 2",
Description3 = "Desc 3",
Description4 = "Desc 4",
Description5 = "Desc 5",
Description6 = "Desc 6",
SerialNumber = "100000",
GroupRecord = GroupRecord,
PriceType = 'I',
Price = 7999.99m,
Year = 1954
};
machineRecords.Create(MachineRecord);
}
EDIT 3:
Now it’s working, this is my working code (for those who are struggling on this too)
Migrations.cs
public class Migrations : DataMigrationImpl
{
public int Create()
{
SchemaBuilder.CreateTable("MachineRecord", table => table
.Column<int>("Id", column => column.PrimaryKey().Identity())
.Column<int>("GroupRecord_Id")
.Column<int>("MachineNumber", column => column.NotNull())
.Column<string>("Title", column => column.NotNull().WithLength(40))
.Column<string>("Description", column => column.WithLength(70))
.Column<char>("PriceType", column => column.NotNull().WithLength(1))
.Column<decimal>("Price", column => column.NotNull())
.Column<int>("Year", column => column.WithLength(4))
);
SchemaBuilder.CreateTable("GroupRecord", table => table
.Column<int>("Id", column => column.PrimaryKey().Identity())
.Column<string>("Name")
);
return 1;
}
}
MachineRecord.cs
public class MachineRecord
{
public virtual int Id { get; set; }
public virtual int MachineNumber { get; set; }
public virtual string Title { get; set; }
public virtual string Description { get; set; }
public virtual GroupRecord GroupRecord { get; set; }
public virtual char PriceType { get; set; }
public virtual decimal Price { get; set; }
public virtual int Year { get; set; }
}
GroupRecord.cs
public class GroupRecord
{
public virtual int Id { get; set; }
public virtual string Name { get; set; }
}
The key to your problem is split across how you structure the models and how you define their relationship in the Migrations.cs.
Here is a solution that I have used in the past.
Firstly, the class which contains the link to the second class.
Note: inheriting from
ContentPartRecordgives you an ID.Then the second class (in this case I do not need to inherit from
ContentPartRecordso I add the ID myself)In the Migrations.cs add the OptionRecord ID in the CaptureRecord table definition
Hopefully this will give you a push in the right direction in terms of how to structure the relationship between your two tables.
Edit: To add content to this sort of structure you will need to manually manage this yourself using the
IRepositoryinstances.So in your controller
AddDummyDatagives an example of this.