bug #21574 [Serializer] Add missing json parameter (javenlalla)

This PR was merged into the 6.4 branch.

Discussion
----------

[Serializer] Add missing json parameter

Hello!
In the `Deserializing Input Partially (Unwrapping)` section of the `Serializer` documentation, the code sample is missing the `'json'` format specifier as the third argument when calling `deserialize`.

This commit adds the missing format value into the function call of the code sample.

Function signature: https://github.com/symfony/symfony/blob/6.4/src/Symfony/Component/Serializer/Serializer.php#L146

Page: https://symfony.com/doc/6.4/serializer.html#deserializing-input-partially-unwrapping

Thanks!

Commits
-------

df65e9f [Serializer] Add missing json parameter

Author: OskarStark

serializer.rst

@@ -2097,7 +2097,7 @@ To avoid deserializing the whole response, you can use the
 and "unwrap" the input data::
 
     $jsonData = '{"result":"success","data":{"person":{"name": "Jane Doe","age":57}}}';
-    $data = $serialiser->deserialize($jsonData, Object::class, [
+    $data = $serialiser->deserialize($jsonData, Object::class, 'json', [
         UnwrappingDenormalizer::UNWRAP_PATH => '[data][person]',
     ]);
     // $data is Person(name: 'Jane Doe', age: 57)