newtonsoft json仅解析16个有效数字

Question

When I parse following json:

{
    "Item1": 123456789.0123456789,
    "Item2": "123456789.0123456789",
    "Item3": 1.234567890123456789,
    "Item4": 1234567890123456789
}

like so:

string data = "{\"Item1\":123456789.0123456789,\"Item2\":\"123456789.0123456789\",\"Item3\":1.234567890123456789,\"Item4\":1234567890123456789}";
JObject json = JsonConvert.DeserializeObject<JObject>(data, new JsonSerializerSettings { DateParseHandling = DateParseHandling.None });
Console.WriteLine(json["Item1"].Value<decimal>());
Console.WriteLine(json["Item2"].Value<decimal>());
Console.WriteLine(json["Item3"].Value<decimal>());
Console.WriteLine(json["Item4"].Value<decimal>());

then I get following output:

123456789,012346
123456789,0123456789
1,23456789012346
1234567890123456789

in my actual json data source inputs are like "name":123456789.0123456789 . How do I parse its whole decimal value without losing precision?

Answer 1

You can use the FloatParseHandling Enumeration to instruct it to deserialize to decimal :

JObject json = JsonConvert.DeserializeObject<JObject>(data, 
    new JsonSerializerSettings { 
        DateParseHandling = DateParseHandling.None, 
        FloatParseHandling = FloatParseHandling.Decimal // Added new setting
    });

(You don't specify which version of Newtonsoft you're using so I assume the latest)