How to create JSON string in C#

I've found that you don't need the serializer at all. If you return the object as a List<type>.
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}]}

Using [Newtonsoft.Json][1] makes it really easier: <br/>

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][2]**


[1]:

[To see links please register here]

[2]:

[To see links please register here]

***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"
]
]
]

Include:

using System.Text.Json;

Then serialize your object_to_serialize like this:
JsonSerializer.Serialize(object_to_serialize)

Simlpe use of **Newtonsoft.Json** and **Newtonsoft.Json.Linq** libraries.

//Create my object
var myData = new
{
Host = @"sftp.myhost.gr",
UserName = "my_username",
Password = "my_password",
SourceDir = "/export/zip/mypath/",
FileName = "my_file.zip"
};

//Tranform it to Json object
string jsonData = JsonConvert.SerializeObject(myData);

//Print the Json object
Console.WriteLine(jsonData);

//Parse the json object
JObject jsonObject = JObject.Parse(jsonData);

//Print the parsed Json object
Console.WriteLine((string)jsonObject["Host"]);
Console.WriteLine((string)jsonObject["UserName"]);
Console.WriteLine((string)jsonObject["Password"]);
Console.WriteLine((string)jsonObject["SourceDir"]);
Console.WriteLine((string)jsonObject["FileName"]);

If you want to avoid creating a class and create JSON then Create a dynamic Object and Serialize Object.

```
dynamic data = new ExpandoObject();
data.name = "kushal";
data.isActive = true;

// convert to JSON
string json = Newtonsoft.Json.JsonConvert.SerializeObject(data);

```
Read the JSON and deserialize like this:
```
// convert back to Object
dynamic output = Newtonsoft.Json.JsonConvert.DeserializeObject(json);

// read a particular value:
output.name.Value
```

`ExpandoObject` is from `System.Dynamic` namespace.