I have a string that comes out of a database which is in Json format.
I have tried to deserialize it with:
RestSharp.Deserializers.JsonDeserializer deserial = new JsonDeserializer(); var x = deserial .Deserialize<Customer>(myStringFromDB)
But the .Deserialize
function expects an IRestResponse
Is there a way to use RestSharp to just deserialize raw strings?
Newtonsoft. Json package is not provided by RestSharp, is marked as obsolete on NuGet, and no longer supported by its creator.
Json structure is made up with {}, [], comma, colon and double quotation marks and it includes the following data types: Object, Number, Boolean, String, and Array. Serialize means convert an object instance to an XML document. Deserialize means convert an XML document into an object instance.
DeserializeObject Method. Deserializes the JSON to a . NET object.
There are sereval ways to do this. A very popular library to handle json is the Newtonsoft.Json
. Probably you already have it on your asp.net project but if not, you could add it from nuget
.
Considering you have a response object, include the following namespaces and call the static method DeserializeObject<T>
from JsonConvert
class:
using Newtonsoft.Json; using RestSharp;
return JsonConvert.DeserializeObject<T>(response.Content);
On the response.Content
, you will have the raw result, so just deserialize this string to a json object. The T
in the case is the type you need to deserialize.
For example:
var customerDto = JsonConvert.DeserializeObject<CustomerDto>(response.Content);
Update
Recently, Microsoft has added a namespace System.Text.Json
which handle json format on the .Net platform. You could use it calling the JsonSerializer.Deserialize<T>
static method:
using System.Text.Json;
var customer = JsonSerializer.Deserialize<Customer>(jsonContent);
If you want to avoid using extra libraries, try this:
RestSharp.RestResponse response = new RestSharp.RestResponse(); response.Content = myStringFromDB; RestSharp.Deserializers.JsonDeserializer deserial = new JsonDeserializer(); Customer x = deserial.Deserialize<Customer>(response);
Caveats apply - not extensively tested - but seems to work well enough.
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