Deserialize json in a “TryParse” way

Question

When I send a request to a service (that I do not own), it may respond either with the JSON data requested, or with an error that looks like this:

{
    \"er         


        

Answer 1

A slightly modified version of @Yuval's answer.

static T TryParse<T>(string jsonData) where T : new()
{
  JSchemaGenerator generator = new JSchemaGenerator();
  JSchema parsedSchema = generator.Generate(typeof(T));
  JObject jObject = JObject.Parse(jsonData);

  return jObject.IsValid(parsedSchema) ?
      JsonConvert.DeserializeObject<T>(jsonData) : default(T);
}

This can be used when you don't have the schema as text readily available for any type.

Answer 2

You may deserialize JSON to a dynamic, and check whether the root element is error. Note that you probably don't have to check for the presence of status and code, like you actually do, unless the server also sends valid non-error responses inside a error node.

Aside that, I don't think you can do better than a try/catch.

What actually stinks is that the server sends an HTTP 200 to indicate an error. try/catch appears simply as checking of inputs.

Answer 3

Just to provide an example of the try/catch approach (it may be useful to somebody).

public static bool TryParseJson<T>(this string obj, out T result)
{
    try
    {
        // Validate missing fields of object
        JsonSerializerSettings settings = new JsonSerializerSettings();
        settings.MissingMemberHandling = MissingMemberHandling.Error;

        result = JsonConvert.DeserializeObject<T>(obj, settings);
        return true;
    }
    catch (Exception)
    {
        result = default(T);
        return false;
    }
}

Then, it can be used like this:

var result = default(MyObject);
bool isValidObject = jsonString.TryParseJson<MyObject>(out result);

if(isValidObject)
{
    // Do something
}

Answer 4

With Json.NET you can validate your json against a schema:

 string schemaJson = @"{
 'status': {'type': 'string'},
 'error': {'type': 'string'},
 'code': {'type': 'string'}
}";

JsonSchema schema = JsonSchema.Parse(schemaJson);

JObject jobj = JObject.Parse(yourJsonHere);
if (jobj.IsValid(schema))
{
    // Do stuff
}

And then use that inside a TryParse method.

public static T TryParseJson<T>(this string json, string schema) where T : new()
{
    JsonSchema parsedSchema = JsonSchema.Parse(schema);
    JObject jObject = JObject.Parse(json);

    return jObject.IsValid(parsedSchema) ? 
        JsonConvert.DeserializeObject<T>(json) : default(T);
}

Then do:

var myType = myJsonString.TryParseJson<AwsomeType>(schema);

Update:

Please note that schema validation is no longer part of the main Newtonsoft.Json package, you'll need to add the Newtonsoft.Json.Schema package.

Update 2:

As noted in the comments, "JSONSchema" have a pricing model, meaning it isn't free. You can find all the information here

Answer 5

@Victor LG's answer using Newtonsoft is close, but it doesn't technically avoid the a catch as the original poster requested. It just moves it elsewhere. Also, though it creates a settings instance to enable catching missing members, those settings aren't passed to the DeserializeObject call so they are actually ignored.

Here's a "catch free" version of his extension method that also includes the missing members flag. The key to avoiding the catch is setting the Error property of the settings object to a lambda which then sets a flag to indicate failure and clears the error so it doesn't cause an exception.

 public static bool TryParseJson<T>(this string @this, out T result)
 {
    bool success = true;
    var settings = new JsonSerializerSettings
    {
        Error = (sender, args) => { success = false; args.ErrorContext.Handled = true; },
        MissingMemberHandling = MissingMemberHandling.Error
    };
    result = JsonConvert.DeserializeObject<T>(@this, settings);
    return success;
}

Here's an example to use it:

if(value.TryParseJson(out MyType result))
{ 
    // Do something with result…
}

Answer 6

To test whether a text is valid JSON regardless of schema, you could also do a check on the number of quotation marks:" in your string response, as shown below :

// Invalid JSON
var responseContent = "asgdg"; 
// var responseContent = "{ \"ip\" = \"11.161.195.10\" }";

// Valid JSON, uncomment to test these
// var responseContent = "{ \"ip\": \"11.161.195.10\", \"city\": \"York\",  \"region\": \"Ontartio\",  \"country\": \"IN\",  \"loc\": \"-43.7334,79.3329\",  \"postal\": \"M1C\",  \"org\": \"AS577 Bell Afgh\",  \"readme\": \"https://ipinfo.io/missingauth\"}";
// var responseContent = "\"asfasf\"";
// var responseContent = "{}";

int count = 0;
foreach (char c in responseContent)
    if (c == '\"') count++; // Escape character needed to display quotation
if (count >= 2 || responseContent == "{}") 
{
    // Valid Json
    try {
        JToken parsedJson = JToken.Parse(responseContent);
        Console.WriteLine("RESPONSE: Json- " + parsedJson.ToString(Formatting.Indented));
    }  
    catch(Exception ex){
        Console.WriteLine("RESPONSE: InvalidJson- " + responseContent);
    }
}
else
    Console.WriteLine("RESPONSE: InvalidJson- " + responseContent);