Howto.md

Introduction (as of v2.1.21)

The interface

To use fastJSON you can use the JSON static object with the following interface :

string ToNiceJSON(object obj)
string ToNiceJSON(object obj, JSONParameters param)
string ToJSON(object obj)
string ToJSON(object obj, JSONParameters param)
object Parse(string json)
dynamic ToDynamic(string json)
T ToObject<T>(string json)
T ToObject<T>(string json, JSONParameters param)
object ToObject(string json)
object ToObject(string json, JSONParameters param)
object ToObject(string json, Type type)
object FillObject(object input, string json)
object DeepCopy(object obj)
T DeepCopy<T>(T obj)
string Beautify(string input)
void RegisterCustomType(Type type, Serialize serializer, Deserialize deserializer)
void ClearReflectionCache()

Basics

Some basic serialization:

// an integer
string s = JSON.ToJSON(42); // s => "42"
// a null
s = JSON.ToJSON(null); // s => null
// a string
s = JSON.ToJSON("hello world!"); // s => "hello world!"
// a date
s = JSON.ToJSON(DateTime.Now()); // s => "2016-07-23T10:40:35Z"

Some basic deserialization:

string s = "\"42\""; // s => the string "42"
var o = JSON.ToObject(s); // o will be a string => "42"
var i = JSON.ToObject<int>(s); // i will be an integer => 42 
var n = JSON.ToObject("null"); // n will be a null object

From the above example immediately you should be aware that since json is essentially a string representation, if you don't specify the type or fastJSON cannot determine the type it will default to a string.

Pretty Print

To get nice formatted json output you can do:

var s = JSON.ToNiceJSON(o);

Custom Types

If fastJSON does not support your types, you can define you own custom type handler with:

// use the following code once in your startup 
JSON.RegisterCustomType(typeof(System.Net.IPAddress),
                        (x) => { return x.ToString(); },
                        (x) => { return System.Net.IPAddress.Parse(x); });

var ip = System.Net.IPAddress.Loopback;
var s = JSON.ToJSON(ip);
var o = JSON.ToObject<System.Net.IPAddress>(s);

Filling existing objects

You can fill an existing object structure with data from a json file which useful for configuration objects etc.

Any values in the json file will overwrite the values in the object structure.

public class config
{
  public int v1;
  public int v2;
}

config c = new config();
c.v1 = 1;
c.v2 = 2;
JSON.FillObject(c, "{\"v2\":10}");
// c.v2 == 10 now

Deep Copy

Sometimes you need to create a copy of something so you can rollback of things go wrong for example, in these situations you can create a copy of the original object by:

var copy = JSON.DeepCopy(obj);
// 'copy' is now the replica of 'obj' and you can change 'obj' without effecting 'copy'
// 'copy' has a different hash code to 'obj'

JSON from other sources

Sometimes you may encounter json from other sources than fastJSON where the type information is absent, in these situations you can do:

public class myType
{
  public string Name;
  public int Age;
}
var str = "{\"Name\":\"bob\",\"age\":42}";
// the json string does not contain fastJSON type information
// so we use ToObject<T> and supply the type we want.
var o = JSON.ToObject<myType>(str);
// also we can use the following if the type can be dynamic
Type t = typeof(myType);
var obj = JSON.ToObject(str, t);

Dynamic objects (.net 4+)

Dynamic objects are used when you don't have the class structure before hand, or you just want to test dynamically generated json data. So given the following json :

{
  "Name" : "Alice",
  "Age" : 42,
  "Address" : "here"
}

You can do :

var obj = JSON.ToDynamic(json);
var name = obj.Name;
var age = obj.Age;
// will give an error since address <> Address in the json 
var address = obj.address; 

Also you can enumerate the following json :

[
  {"Name":"Alice"},
  {"Name":"Bob"},
  {"Name":"Carol"}
]

With the following code:

foreach(var o in JSON.ToDynamic(json))
{
    Console.WriteLine(o.Name);    
}