コード例 #1
0
        public static ITableInfoBuilder Named(this ITableInfoBuilder builder, ObjectName tableName)
        {
            if (tableName == null)
            {
                throw new ArgumentNullException("tableName");
            }

            return(builder.InSchema(tableName.ParentName).Named(tableName.Name));
        }
コード例 #2
0
 public static ITableInfoBuilder WithColumn(this ITableInfoBuilder builder, string columnName, SqlType columnType)
 {
     return(builder.WithColumn(column => column.Named(columnName).HavingType(columnType)));
 }