C#•7mo ago

✅ @ in the property name c#

Team , Is there any way to get @ in the property name like this public class TestClass { [JsonProperty("@context)] public string @context { get; set; } } Adding Jsonproperty , I am able to deserialize . but I need @context as output property,right now it is coming as just context But Underscore is working public class TestClass { [JsonProperty("@context)] public string _context { get; set; } } Please guide on this

24 Replies

Angius•7mo ago

Why would you want the property to be called that, and not Context? [JsonProperty] handles the serialization and deserialization The name in code can just as well be UngaBunga

Shiv•7mo ago

I need to send the same object back to an library which is expecting an @

Angius•7mo ago

It's not a valid identifier, so no library can use it either Not to mention, unless using reflections, the names of properties don't matter

Unknown User•7mo ago

Message Not Public

Angius•7mo ago

Json has a property named @context from what I gather

Unknown User•7mo ago

Message Not Public

Angius•7mo ago

Probably Idk, detail is scarce lol

Shiv•7mo ago

So it is conflicting?

Unknown User•7mo ago

Message Not Public

Angius•7mo ago

What is conflicting with what? Handling a Json property named @context is easy with [JsonProperty] attribute The property itself cannot begin witn @ No library will ever expect a property starting with @ Why do you need specifically the property of your class to be named @Context and not Context or UngaBunga?

Shiv•7mo ago

It is an json LD object , if i send without @ it is throwing an error saying invalid object https://json-ld.org/

Angius•7mo ago

[JsonProperty] handles the conversion to Json

MODiX•7mo ago

Angius

REPL Result: Success

using System.Text.Json;

public class Foo
{
  [JsonProperty("@context")]
  public string UngaBunga { get; set; }
}

var foo = new Foo { UngaBunga = "hello world" };

var json = JsonSerializer.Serialize(foo);

Console.WriteLine(json);

var newFoo = JsonSerializer.Deserialize<Foo>(json);

return newFoo;

using System.Text.Json;

public class Foo
{
  [JsonProperty("@context")]
  public string UngaBunga { get; set; }
}

var foo = new Foo { UngaBunga = "hello world" };

var json = JsonSerializer.Serialize(foo);

Console.WriteLine(json);

var newFoo = JsonSerializer.Deserialize<Foo>(json);

return newFoo;

Result: Foo

{
  "ungaBunga": "hello world"
}

{
  "ungaBunga": "hello world"
}

Console Output

{"UngaBunga":"hello world"}

{"UngaBunga":"hello world"}

Compile: 660.550ms | Execution: 56.581ms | React with ❌ to remove this embed.

Angius•7mo ago

Huh, what? [JsonProperty] doesn't work for serialization?

Unknown User•7mo ago

Message Not Public

Angius•7mo ago

Ah Well that's the issue lmao

Unknown User•7mo ago

Message Not Public

MODiX•7mo ago

Angius

REPL Result: Success

using System.Text.Json;
using System.Text.Json.Serialization;

public class Foo
{
  [JsonPropertyName("@context")]
  public string UngaBunga { get; set; }
}

var foo = new Foo { UngaBunga = "hello world" };

var json = JsonSerializer.Serialize(foo);

Console.WriteLine(json);

var newFoo = JsonSerializer.Deserialize<Foo>(json);

return newFoo;

using System.Text.Json;
using System.Text.Json.Serialization;

public class Foo
{
  [JsonPropertyName("@context")]
  public string UngaBunga { get; set; }
}

var foo = new Foo { UngaBunga = "hello world" };

var json = JsonSerializer.Serialize(foo);

Console.WriteLine(json);

var newFoo = JsonSerializer.Deserialize<Foo>(json);

return newFoo;

Result: Foo

{
  "@context": "hello world"
}

{
  "@context": "hello world"
}

Console Output

{"@context":"hello world"}

{"@context":"hello world"}

Compile: 657.888ms | Execution: 96.639ms | React with ❌ to remove this embed.

Angius•7mo ago

There

Shiv•7mo ago

Should I use System.Text.Json; and not Newtonsoft.Json; ? @TeBeCo

Unknown User•7mo ago

Message Not Public

Angius•7mo ago

STJ is easier, better, faster, stronger

Angius•7mo ago

https://tenor.com/view/daftpunk-harder-better-faster-stronger-gif-11845009

Tenor

Shiv•7mo ago

Got it. Thanks for the input @ZZZZZZZZZZZZZZZZZZZZZZZZZ @TeBeCo

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✅ @ in the property name c#