使用Umbraco.Core.Persistence从带有枚举的模型创建数据库表

Question

要在应用程序启动时创建此表（如果不存在）。

码：

public class Database : ApplicationEventHandler
{
    protected override void ApplicationStarted(UmbracoApplicationBase umbracoApplication, ApplicationContext applicationContext)
    {
        var db = applicationContext.DatabaseContext.Database;

        //Cant add this table due to the ENUM
        if (!db.TableExist("FormData"))
        {
            db.CreateTable<FormData>(false);
        }
    }
}

模型：

[PrimaryKey("Id")]
public class FormData
{
    [PrimaryKeyColumn(AutoIncrement = true, IdentitySeed = 1)]
    public int Id { get; set; }

    [NullSetting(NullSetting = NullSettings.NotNull)]
    public FormType Type { get; set; }

    [NullSetting(NullSetting = NullSettings.NotNull)]
    public string Data { get; set; }

    [NullSetting(NullSetting = NullSettings.NotNull)]
    public DateTime Date { get; set; }
}

错误信息：

[InvalidOperationException：序列不包含匹配的元素] System.Linq.Enumerable.First（IEnumerable 1 source, Func2谓词）+415 Umbraco.Core.Persistence.SqlSyntax.SqlSyntaxProviderBase 1.FormatType(ColumnDefinition column) +1225 Umbraco.Core.Persistence.SqlSyntax.SqlSyntaxProviderBase1.Format（ColumnDefinition列）+155 Umbraco.Core.Persistence.SqlSyntax.SqlSyntaxProviderBase 1.Format(IEnumerable1列）+144 Umbraco.Core.Persistence.SqlSyntax.SqlSyntaxProviderBase`1.Format（TableDefinition table）+131 Umbraco.Core.Persistence.PetaPocoExtensions.CreateTable（数据库db，布尔覆盖，类型为ModelType）+161 Umbraco.Core.Persistence .PetaPocoExtensions.CreateTable（数据库db，布尔覆盖）+121

看着错误，我认为不更新内核就不会有解决方案，但这是希望你们能提供帮助

Answer 1

在研究@Ryios给出的答案时，我认为类似这样的东西很好：

/// <summary>
/// Don't use this to get or set.
/// This must however be kept as public or db.CreateTable()
/// will not insert this field into the database.
/// </summary>
[NullSetting(NullSetting = NullSettings.NotNull)]
[Column("type")]
public int _type { get; set; }

/// <summary>
/// This field is ignored by db.CreateTable().
/// </summary>
[Ignore]
public FormType Type
{
    get
    {
        return (FormType)_type;
    }
    set
    {
        _type = (int)value;
    }
}

在代码中，Type应该使用而不是_type为了使枚举受益。_type仅作为插入数据库表中的字段出现。