簡體   English   中英

使用代碼優先實體框架格式化序列化JSON

[英]Formatting Serialized JSON With Code-First Entity Framework

我正在使用Web-API GET返回報告的JSON。

GET方法返回一個簡單的db.Reports.ToList();

這是我檢索到的數據的轉儲

{
        "Project": {
            "Location": {
                "LocationId": 7,
                "Description": "New York"
            },
            "Department": {
                "DepartmentId": 7,
                "Description": "Engineering"
            },
            "ProjectId": 7,
            "Description": "Project_3",
            "LocationId": 7,
            "DepartmentId": 7
        },
        "Person": {
            "Email": "email@gmail.com",
            "FirstName": "John",
            "LastName": "Doe",
            "IsActive": true
        },
        "StatusCode": {
            "StatusId": 8,
            "Description": "Accepted"
        },
        "ReportId": "d4cddb3f-ea6a-4b0a-9820-19bd8ee43b3a",
        "Description": "Report 3",
        "RoundTrip": 45.88,
        "IsBillable": true,
        "StartDate": "2013-06-27T00:00:00",
        "EndDate": "2013-06-27T14:36:32.467",
        "TimeUpdated": "AAAAAAAAJxM="
    }, ... 
}

這是相關的報告聲明:

public class Report
{
    public Guid ReportId { get; set; }
    public string Description { get; set; }
    public double RoundTrip { get; set; }
    public bool IsBillable { get; set; }

    public DateTime StartDate { get; set; }
    public DateTime EndDate { get; set; }

    public virtual Project Project { get; set; }

    public byte[] TimeUpdated { get; set; }

    public virtual Person Person { get; set; }

    public virtual StatusCode StatusCode { get; set; }
}

在這種情況下,我真的很想在Report類中包含各種對象的ID。 例如,我真的很想看看:

    "Project": 7,
    "Location": 7,
    "Department": 7,
    "Person": "email@gmail.com",
    "StatusCode": 8,
    "ReportId": "d4cddb3f-ea6a-4b0a-9820-19bd8ee43b3a",
    "Description": "Report 3",
    "RoundTrip": 45.88,
    "IsBillable": true,
    "StartDate": "2013-06-27T00:00:00",
    "EndDate": "2013-06-27T14:36:32.467",
    "TimeUpdated": "AAAAAAAAJxM="
  1. 有沒有相對簡單的方法可以進行此操作,或者只是進一步解析我已經看到的結果是否符合我的利益?
  2. 為什么EF默認情況下會在JSON中創建這些對象,而不僅僅是外鍵?

您無需專門創建一個新類即可。 ApiController ,如果使用類型化的返回類型(支持HttpResponseMessage ),則將List類型更改為IEnumerable<object>並返回:

return db.Reports.Select(r => new {
    Project = r.ProjectId;
    Location = r.Location.LocationId;
    Department = r.Department.DepartmentId;
    Person = r.Person.Email;
    StatusCode = r.StatusCode.StatusId;
    Description: r.Description
    RoundTrip: r.RoundTrip
    IsBillable: r.IsBillable,
    StartDate: r.StartDate,
    EndDate: r.EndDate
    TimeUpdated: r.TimeUpdated
});

// Or if you're using HttpResponseMessage
return Request.CreateResponse(HttpStatusCode.Ok, 
    db.Reports.Select(r => new {
        Project = r.ProjectId;
        Location = r.Location.LocationId;
        Department = r.Department.DepartmentId;
        Person = r.Person.Email;
        StatusCode = r.StatusCode.StatusId;
        Description: r.Description
        RoundTrip: r.RoundTrip
        IsBillable: r.IsBillable,
        StartDate: r.StartDate,
        EndDate: r.EndDate
        TimeUpdated: r.TimeUpdated
    }));

默認的Json序列化程序(Newtonsoft的Json.Net)足夠聰明,可以序列化匿名對象。 上面代碼中唯一未知的是TimeUpdated成員的行為,因為它是一個字節數組。 您可能需要調整分配。

我建議您創建一個用於顯示JSON的模型。 這將是最簡單的選擇。

這樣的事情應該起作用:

public class ReportSimple
{
    public Guid ReportId { get; set; }
    public int Project { get; set; }
    public int Location { get; set; } 
    public int Department { get; set; }
    public string Person { get; set; }
    public int StatusCode { get; set; }
    public string Description { get; set; }
    public double RoundTrip { get; set; }
    public bool IsBillable { get; set; }
    public DateTime StartDate { get; set; }
    public DateTime EndDate { get; set; }
    public byte[] TimeUpdated { get; set; }

    public ReportSimple(Project project, Person person, StatusCode statusCode)
    {
        Project = project.ProjectId;
        Location = project.Location.LocationId;
        Department = project.Department.DepartmentId;
        Person = person.Email;
        StatusCode = statusCode.StatusId;
    }
}

暫無
暫無

聲明:本站的技術帖子網頁,遵循CC BY-SA 4.0協議,如果您需要轉載,請注明本站網址或者原文地址。任何問題請咨詢:yoyou2525@163.com.

 
粵ICP備18138465號  © 2020-2024 STACKOOM.COM