Inverse of `.MapTo<'T>`

まず、既存の MapTo<'T'> の使用方法と、issue に示されている Create/SubmitUpdates のフローを確認します。レコードをエンティティに変換するために必要な API とマッピング動作を特定します。これには、列ごとの割り当てなしで例がどのように動作すべきかも含まれます。動作とサポート対象のケースが定義され、検証されれば完了です。

Is your feature request related to a problem? Please describe.
This library provides the .MapTo<'T> to map an entity to a record type (which is awesome), but it is not as simple to map this record type back to an entity. Mapping back can be done manually, but it can get a little cumbersome when the entity has a lot of columns.

Describe the solution you'd like
It would be cool to have some sort of inverse of the .MapTo<'T> method that could take a record and return an entity.

Describe alternatives you've considered
As mentioned above, it can currently be done manually. This works fine for small entities, but scales with the number of columns in the entity.

Basic Example
If we have a table MyTable with column Id that is mapped by the record MyTable_Row, you can easily grab a row based on the column Id using

let getById (id: int): MyTable_Row option =
    let foundRow = query {
        for row in ctx.Dbo.MyTable do
        where (row.Id = id)
        select (Some row)
        exactlyOneOrDefault
    }
    match foundRow with
    | Some foundRow -> Some( foundRow.MapTo<MyTable_Row>() )
    | None          -> None

This method is independent of the number of columns in MyTable (other than in the definition of MyTable_Row). But, if we want to add a row to MyTable from a MyTable_Row, we need to manually specify that each column gets populated:

let insertRow (myRow: MyTable_Row) =
    let newRow = ctx.Dbo.MyTable.Create()
    newRow.Id      <- myRow.Id
    newRow.Column1 <- myRow.Column1
    newRow.Column2 <- myRow.Column2
    newRow.Column3 <- myRow.Column3
    newRow.Column4 <- myRow.Column4
    ...
    newRow.ColumnN <- myRow.ColumnN
    ctx.SubmitUpdates()

With this, not only does the definition of the record type need to scale with the number of columns in the entity, but all functions performing inserts must too.