ServiceStack DTO映射问题

Question

我们在4.0.56中看到了我们之前见过的问题（参见ServiceStack - [Reference] or [Ignore]?的更新3） - 即，正在使用TitleID值填充员工的ID属性。

请求DTO是：

[Route("/employees", "GET")]
public class FindEmployeesRequest : QueryDb<Employee>,
    IJoin<Employee, EmployeeType>,
    IJoin<Employee, Department>,
    IJoin<Employee, Title> {

    public int? ID { get; set; }
    public int[] IDs { get; set; }
    public string UserID { get; set; }
    public string[] UserIDs { get; set; }
    public int? EmployeeTypeID { get; set; }
    public int[] EmployeeTypeIDs { get; set; }
    public int? DepartmentID { get; set; }
    public int[] DepartmentIDs { get; set; }
    public int? TitleID { get; set; }
    public int[] TitleIDs { get; set; }
    public string LastNameStartsWith { get; set; }
    public DateTime[] DateOfBirthBetween { get; set; }
    public DateTime[] HireDateBetween { get; set; }
    public bool? IsActive { get; set; }
    public bool? IsEligibleToOrderLunch { get; set; }

    [QueryDbField(Template = "(MONTH({Field}) = {Value})", Field = "DateOfBirth")]
    public int? BirthMonth { get; set; }

    [QueryDbField(Template = "(DAY({Field}) = {Value})", Field = "DateOfBirth")]
    public int? BirthDay { get; set; }

    [QueryDbField(Template = "({Field} LIKE '%(Incoming)')", Field = "EmployeeTypeName")]
    public bool? IsIncoming { get; set; }

    [QueryDbField(Template = "(FirstName LIKE {Value} OR LastName LIKE {Value} OR PreferredName LIKE {Value})", ValueFormat = "%{0}%", Field = "ID")]
    public string NameSearch { get; set; }

    [QueryDbField(Template = "(FirstName LIKE {Value} OR LastName LIKE {Value} OR PreferredName LIKE {Value} OR Department.Name LIKE {Value} OR Title.Name LIKE {Value})", ValueFormat = "%{0}%", Field = "ID")]
    public string BasicSearch { get; set; }

    [QueryDbField(Template = "({Field} LIKE {Value})", Field = "EmployeeTypeName", ValueFormat = "%{0}%")]
    public string EmployeeTypeSearch { get; set; }

    [QueryDbField(Template = "({Field} LIKE {Value})", Field = "DepartmentName", ValueFormat = "%{0}%")]
    public string DepartmentSearch { get; set; }

    [QueryDbField(Template = "({Field} LIKE {Value})", Field = "TitleName", ValueFormat = "%{0}%")]
    public string TitleSearch { get; set; }
}

基本上任何执行＆＃34; SELECT * FROM＆＃34;的请求展示了这个问题 - 这里有一些来自Profiler的SQL＆＃34; / employees？id = 101＆amp; fields = *＆＃34;：

exec sp_executesql N'SELECT * 
FROM "Employee" INNER JOIN "EmployeeType" ON
("EmployeeType"."ID" = "Employee"."EmployeeTypeID") INNER JOIN "Department" ON
("Department"."ID" = "Employee"."DepartmentID") INNER JOIN "Title" ON
("Title"."ID" = "Employee"."TitleID")
WHERE "Employee"."ID" = @0',N'@0 int',@0=101

当我自己运行此查询时，＆＃34;员工＆＃34;。＆＃34; ID＆＃34;是101，但在服务的回复中，它是67（TitleID）：

Answer 1

您不能自己使用裸露的通配符?fields=*，它需要包含表格前缀，例如?fields=employee.*为mentioned in the docs，这是手动指定表格中每个字段的简写。