How to create JSON string in C#

问题:

I just used the XmlWriter to create some XML to send back in an HTTP response. How would you create a JSON string. I assume you would just use a stringbuilder to build the JSON string and them format your response as JSON?

回答1:

You could use the JavaScriptSerializer class, check this article to build an useful extension method.

Code from article:

namespace ExtensionMethods
{
    public static class JSONHelper
    {
        public static string ToJSON(this object obj)
        {
            JavaScriptSerializer serializer = new JavaScriptSerializer();
            return serializer.Serialize(obj);
        }

        public static string ToJSON(this object obj, int recursionDepth)
        {
            JavaScriptSerializer serializer = new JavaScriptSerializer();
            serializer.RecursionLimit = recursionDepth;
            return serializer.Serialize(obj);
        }
    }
}

Usage:

using ExtensionMethods;

...

List<Person> people = new List<Person>{
                   new Person{ID = 1, FirstName = \"Scott\", LastName = \"Gurthie\"},
                   new Person{ID = 2, FirstName = \"Bill\", LastName = \"Gates\"}
                   };


string jsonString = people.ToJSON();

回答2:

Using Newtonsoft.Json makes it really easier:

Product product = new Product();
product.Name = \"Apple\";
product.Expiry = new DateTime(2008, 12, 28);
product.Price = 3.99M;
product.Sizes = new string[] { \"Small\", \"Medium\", \"Large\" };

string json = JsonConvert.SerializeObject(product);

Documentation: Serializing and Deserializing JSON

回答3:

This library is very good for JSON from C#

http://james.newtonking.com/pages/json-net.aspx

回答4:

This code snippet uses the DataContractJsonSerializer from System.Runtime.Serialization.Json in .NET 3.5.

public static string ToJson<T>(/* this */ T value, Encoding encoding)
{
    var serializer = new DataContractJsonSerializer(typeof(T));

    using (var stream = new MemoryStream())
    {
        using (var writer = JsonReaderWriterFactory.CreateJsonWriter(stream, encoding))
        {
            serializer.WriteObject(writer, value);
        }

        return encoding.GetString(stream.ToArray());
    }
}

回答5:

Take a look at http://www.codeplex.com/json/ for the json-net.aspx project. Why re-invent the wheel?

回答6:

You can also try my ServiceStack JsonSerializer it\'s the fastest .NET JSON serializer at the moment. It supports serializing DataContracts, any POCO Type, Interfaces, Late-bound objects including anonymous types, etc.

Basic Example

var customer = new Customer { Name=\"Joe Bloggs\", Age=31 };
var json = JsonSerializer.SerializeToString(customer);
var fromJson = JsonSerializer.DeserializeFromString<Customer>(json); 

Note: Only use Microsofts JavaScriptSerializer if performance is not important to you as I\'ve had to leave it out of my benchmarks since its up to 40x-100x slower than the other JSON serializers.

回答7:

If you need complex result (embedded) create your own structure:

class templateRequest
{
    public String[] registration_ids;
    public Data data;
    public class Data
    {
        public String message;
        public String tickerText;
        public String contentTitle;
        public Data(String message, String tickerText, string contentTitle)
        {
            this.message = message;
            this.tickerText = tickerText;
            this.contentTitle = contentTitle;
        }                
    };
}

and then you can obtain JSON string with calling

List<String> ids = new List<string>() { \"id1\", \"id2\" };
templateRequest request = new templeteRequest();
request.registration_ids = ids.ToArray();
request.data = new templateRequest.Data(\"Your message\", \"Your ticker\", \"Your content\");

string json = new JavaScriptSerializer().Serialize(request);

The result will be like this:

json = \"{\\\"registration_ids\\\":[\\\"id1\\\",\\\"id2\\\"],\\\"data\\\":{\\\"message\\\":\\\"Your message\\\",\\\"tickerText\\\":\\\"Your ticket\\\",\\\"contentTitle\\\":\\\"Your content\\\"}}\"

Hope it helps!

回答8:

If you can\'t or don\'t want to use the two built-in JSON serializers (JavaScriptSerializer and DataContractJsonSerializer) you can try the JsonExSerializer library - I use it in a number of projects and works quite well.

回答9:

If you\'re trying to create a web service to serve data over JSON to a web page, consider using the ASP.NET Ajax toolkit:

http://www.asp.net/learn/ajax/tutorial-05-cs.aspx

It will automatically convert your objects served over a webservice to json, and create the proxy class that you can use to connect to it.

回答10:

The DataContractJSONSerializer will do everything for you with the same easy as the XMLSerializer. Its trivial to use this in a web app. If you are using WCF, you can specify its use with an attribute. The DataContractSerializer family is also very fast.

回答11:

I\'ve found that you don\'t need the serializer at all. If you return the object as a List. Let me use an example.

In our asmx we get the data using the variable we passed along

// return data
[WebMethod(CacheDuration = 180)]
public List<latlon> GetData(int id) 
{
    var data = from p in db.property 
               where p.id == id 
               select new latlon
               {
                   lat = p.lat,
                   lon = p.lon

               };
    return data.ToList();
}

public class latlon
{
    public string lat { get; set; }
    public string lon { get; set; }
}

Then using jquery we access the service, passing along that variable.

// get latlon
function getlatlon(propertyid) {
var mydata;

$.ajax({
    url: \"getData.asmx/GetLatLon\",
    type: \"POST\",
    data: \"{\'id\': \'\" + propertyid + \"\'}\",
    async: false,
    contentType: \"application/json;\",
    dataType: \"json\",
    success: function (data, textStatus, jqXHR) { //
        mydata = data;
    },
    error: function (xmlHttpRequest, textStatus, errorThrown) {
        console.log(xmlHttpRequest.responseText);
        console.log(textStatus);
        console.log(errorThrown);
    }
});
return mydata;
}

// call the function with your data
latlondata = getlatlon(id);

And we get our response.

{\"d\":[{\"__type\":\"MapData+latlon\",\"lat\":\"40.7031420\",\"lon\":\"-80.6047970}]}

回答12:

Encode Usage

Simple object to JSON Array EncodeJsObjectArray()

public class dummyObject
{
    public string fake { get; set; }
    public int id { get; set; }

    public dummyObject()
    {
        fake = \"dummy\";
        id = 5;
    }

    public override string ToString()
    {
        StringBuilder sb = new StringBuilder();
        sb.Append(\'[\');
        sb.Append(id);
        sb.Append(\',\');
        sb.Append(JSONEncoders.EncodeJsString(fake));
        sb.Append(\']\');

        return sb.ToString();
    }
}

dummyObject[] dummys = new dummyObject[2];
dummys[0] = new dummyObject();
dummys[1] = new dummyObject();

dummys[0].fake = \"mike\";
dummys[0].id = 29;

string result = JSONEncoders.EncodeJsObjectArray(dummys);

Result: [[29,\"mike\"],[5,\"dummy\"]]

Pretty Usage

Pretty print JSON Array PrettyPrintJson() string extension method

string input = \"[14,4,[14,\\\"data\\\"],[[5,\\\"10.186.122.15\\\"],[6,\\\"10.186.122.16\\\"]]]\";
string result = input.PrettyPrintJson();

Results is:

[
   14,
   4,
   [
      14,
      \"data\"
   ],
   [
      [
         5,
         \"10.186.122.15\"
      ],
      [
         6,
         \"10.186.122.16\"
      ]
   ]
]