How to use Json File to serve as a data source for webapi

Question

I want to retrieve data from Json file with webapi using Json file as data source.

In the controller, I have added following code:

Public HttpResponseMessage Get()
{
    var json = File.ReadAllText(System.Web.HttpContext.Current.Server.MapPath(@"~/App_Data\Bim.Json"));

      return new HttpResponseMessage()
    {
        Content = new StringContent(json, Encoding.UTF8, "application/json"),
        StatusCode = HttpStatusCode.OK
    };
}

and in model,

public class bimModel
{
    public string name { get; set; }
    public string material { get; set; }
    public string tag { get; set; }
    public string contentType { get; set; }
}

Actually, I want to use a json file which is located in app_data folder, how to read data from JSON file? I am very new to the webApi, so it will be of great help if the experts can post the entire working code or as many details as possible, please.

your question is not clear, what is your issue? why does your Action not work? why do you have a model where is it used? — Victor Hugo Terceros
– Victor Hugo Terceros, Commented Mar 20, 2018 at 21:33

marc_s · Accepted Answer · 2019-12-11 19:02:57Z

2

If I am correct, I believe you want to deserialize the json text file to model classes, then serialize them back to json for the client.

To do this, you can use https://www.nuget.org/packages/newtonsoft.json/ to help you out. Follow the instructions to get the package installed.

Once you do that, you can write a function such as

public bimModel Get()
{
    var json = File.ReadAllText(System.Web.HttpContext.Current.Server.MapPath(@"~/App_Data\Bim.Json"));
    var model = JsonConvert.DeserializeObject<bimModel>(json);

    return model;
}

If your json file is an array of bimModels, then you can change the return type and type parameter of DeserializeObject to List<bimModel>.

To ensure your server is returning json to the client, make sure your WebApiConfig class is using the JsonMediaTypeFormatter

public static void Register(HttpConfiguration config)
{
    config.MapHttpAttributeRoutes();
    config.Formatters.Clear();
    config.Formatters.Add(new JsonMediaTypeFormatter());
    config.Formatters.JsonFormatter.SerializerSettings.ContractResolver = new CamelCasePropertyNamesContractResolver();
    config.Formatters.JsonFormatter.SerializerSettings.NullValueHandling = NullValueHandling.Ignore;
    config.Formatters.JsonFormatter.SerializerSettings.Formatting = Formatting.Indented;
}

And of course, be sure to include using Newtonsoft.Json; and using Newtonsoft.Json.Serialization;

edited Dec 11, 2019 at 19:02

marc_s

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answered Mar 21, 2018 at 1:20

Kevin Le

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2 Comments

Kumar Over a year ago

Thank you @Kevin Le

Kevin Le Over a year ago

If this works, would you mind upvoting/marking as answered? Thanks!

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