JSON

using System.Text.Json;
using System.Text.Json.Serialization;
using System.Text.Encodings.Web;
using System.Text.Json.Nodes;

JSON is often used to serialize (convert to string) and deserialize (convert from string) various objects (instances of classes and structs). Let's create an object. The class is at the end of this recipe.

var x = new Example {
	Property = 100,
	p = new(10, 20),
	a = ["one", "two"],
};

Will need a JsonSerializerOptions. Important for speed: use the same options variable for all operations that use the same options.

JsonSerializerOptions options = new () {
	DefaultIgnoreCondition = JsonIgnoreCondition.WhenWritingNull,
	IncludeFields = true,
	IgnoreReadOnlyFields = true,
	IgnoreReadOnlyProperties = true,
	WriteIndented = true,
	Encoder = JavaScriptEncoder.UnsafeRelaxedJsonEscaping,
	AllowTrailingCommas = true,
};

Serialize the object with JsonSerializer.

string s = JsonSerializer.Serialize<Example>(x, options); //convert to JSON string
print.it(s);
//File.WriteAllBytes(@"C:\Test\test.json", JsonSerializer.SerializeToUtf8Bytes<Example>(v, options)); //save in file

Deserialize.

var y = JsonSerializer.Deserialize<Example>(s, options); //convert from JSON string
print.it(y);
//var y = JsonSerializer.Deserialize<Example>( //load file
//	File.ReadAllBytes(@"C:\Test\test.json"),
//	options);

To get values from a JSON string without converting it to an object, use class JsonNode. See recipe Http POST. The class also can be used to create a JSON string. Look for more info on the internet.

var j = JsonNode.Parse(File.ReadAllBytes(@"C:\Test\httpPost.json"));
print.it(j["headers"]["Host"]);
foreach (var v in j.AsObject()) {
	print.it(v);
}

JSON is a good format for various settings. The JSettings class uses it. See recipe saving variables.

An example class.

record class Example {
	public int Property { get; set; }
	public string field = "text";
	public POINT p;
	public string[] a;
	
	[JsonIgnore]
	public int excludedExplicitly = 1;
	int _excludedBecausePrivate = 2;
}