Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
554 views
in Technique[技术] by (71.8m points)

c# - Linq query JObject

I am using Json.net for serializing and then making an JObject that looks like this:

 "RegistrationList": [
    {
      "CaseNumber": "120654-1330",
      "Priority": 5,
      "PersonId": 7,
      "Person": {
        "FirstName": "",
        "LastName": "",
      },
      "UserId": 7,
      "User": {
        "Id": 7,
        "CreatedTime": "2013-07-05T13:09:57.87",
        "Comment": "",
    },

How do i query this into a new Object or list, that is easily put into some html table/view. I only want to display the CaseNumber, FirstName and Comment.

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

I only want to display the CaseNumber, FirstName and Comment.

As always in ASP.NET MVC you could start by writing a view model that matches your requirements:

public class MyViewModel
{
    public string CaseNumber { get; set; }
    public string FirstName { get; set; }
    public string Comment { get; set; }
}

then in your controller action you build the view model from the JObject instance you already have:

public ActionResult Index()
{
    JObject json = ... the JSON shown in your question (after fixing the errors because what is shown in your question is invalid JSON)

    IEnumerable<MyViewModel> model =
        from item in (JArray)json["RegistrationList"]
        select new MyViewModel
        {
            CaseNumber = item["CaseNumber"].Value<string>(),
            FirstName = item["Person"]["FirstName"].Value<string>(),
            Comment = item["User"]["Comment"].Value<string>(),
        };

    return View(model);
}

and finally in your strongly typed view you display the desired information:

@model IEnumerable<MyViewModel>

<table>
    <thead>
        <tr>
            <th>Case number</th>
            <th>First name</th>
            <th>Comment</th>
        </tr>
    </thead>
    <tbody>
        @foreach (var item in Model)
        {
            <tr>
                <td>@item.CaseNumber</td>
                <td>@item.FirstName</td>
                <td>@item.Comment</td>
            </tr>
        }
    </tbody>
</table>

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...