你可以试试这个

JObject education = (JObject)JObject.Parse(json)["data"]["education"][0];

    foreach (var prop in education.Properties())
    {
        if (prop.Value.Type != JTokenType.Object) Console.WriteLine("\r\nName: " + prop.Name + "  Value: " + prop.Value);
        else
        {
            Console.WriteLine("\r\nName: " + prop.Name + " Values:");
            foreach (var p in ((JObject)prop.Value).Properties())
            {
                Console.WriteLine(p.Name + " -- " + p.Value);
            }
        }
    }

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输出量

Name: id  Value: 29364744

Name: organization  Value: Texas A&M University

Name: accreditation Values:
education -- Bachelor of Science
educationLevel -- bachelors
inputStr -- Bachelor of Science Computer Science
matchStr -- 

Name: grade  Value: 

Name: location Values:
formatted -- College Station, TX, USA
streetNumber -- 
street -- 
apartmentNumber -- 
city -- College Station
postalCode -- 
state -- Texas
country -- United States
rawInput -- College Station, TX
countryCode -- US
latitude -- 30.627977
longitude -- -96.3344068

Name: dates Values:
startDate -- 2005-01-01
completionDate -- 2009-01-01
isCurrent -- False
rawText -- 2005 - 2009

型

创建一个具有相关属性的类，并将该类型传递给DeserializeObject。

public class Education
    {
        public string id { get; set; }
        public string organization { get; set; }
    }

    public class Employee
    {
        public List<Education> education { get; set; }
    }

    public class Base
    {
       public Employee data { get; set; }
    }

var response = client.Post(request);
var json = response.Content.ToString();
var data = JsonConvert.DeserializeObject<Base>(json);

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你可以用你的json结构创建一个类visual studio有一个选项

的数据
在使用jsonnet之后，您可以反序列化消息

myObject jsonobj = JsonConvert.DeserializeObject<myObject>(json);

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