将 Rest API JSON 响应转换为 C# 对象

Question

我有一个代码 REST API 响应，它是 json，解析为 JObject 并从中提取一个值。但是在解析为 JObject 时出现错误。

错误：“解析值时遇到意外字符：S。路径 ''，第 0 行，位置 0。”

有没有其他方法可以将 Json 字符串转换为 C# 对象。

我有以下代码：使用 Newtonsoft.Json;

    using (HttpResponseMessage message = httpclient.GetAsync(folderIdURL).Result)
    {
        if(message.IsSuccessStatusCode)
        {
            var dataobjects = message.Content.ReadAsStringAsync();
            //dataobjects = "{"id":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/","title":"DQL query results","author":[{"name":"EMC Documentum"}],"updated":"2019-05-02T15:19:52.508+00:00","page":1,"items-per-page":100,"links":[{"rel":"self","href":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/?dql=SELECT%20r_object_id%2cobject_name%20FROM%20dm_sysobject%20WHERE%20FOLDER%20(%27%2fgbc%2fUS%2fOSA-ATTACHMENT%2f2019%27)"}],"entries":[{"id":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/?dql=SELECT%20r_object_id%2cobject_name%20FROM%20dm_sysobject%20WHERE%20FOLDER%20(%27%2fgbc%2fUS%2fOSA-ATTACHMENT%2f2019%27)&index=0","title":"0b0111738011c114","updated":"2019-05-02T15:19:52.508+00:00","published":"2019-05-02T15:19:52.508+00:00","links":[{"rel":"edit","href":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/objects/0b0111738011c114"}],"content":{"json-root":"query-result","definition":"https://gbc-dev5.cloud.wc.com/DctmRest/repositori                      es/dmgbsap_crt/types/dm_sysobject","properties":{"r_object_id":"0b0111738011c114","object_name":"04"},"links":[{"rel":"self","href":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/objects/0b0111738011c114"}]}},{"id":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/?dql=SELECT%20r_object_id%2cobject_name%20FROM%20dm_sysobject%20WHERE%20FOLDER%20(%27%2fgbc%2fUS%2fOSA-ATTACHMENT%2f2019%27)&index=1","title":"0b0111738011c115","updated":"2019-05-02T15:19:52.509+00:00","published":"2019-05-02T15:19:52.509+00:00","links":[{"rel":"edit","href":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/objects/0b0111738011c115"}],"content":{"json-root":"query-result","definition":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/types/dm_sysobject","properties":{"r_object_id":"0b0111738011c115","object_name":"05"},"links":[{"rel":"self","href":"https://gbc-dev5.cloud.wc.com/DctmRest/repositories/dmgbsap_crt/objects/0b0111738011c115"}]}}]}"

            JObject responseObj = JObject.Parse(dataobjects.ToString());
            String id = (String)responseObj["entries" -->"content"-->"properties"-->"object_name"];
        }                                      

    }

}

我期待来自 (String)responseObject["enteries"]["content"]["properties"]["object_name"] 的值

Answer 1

JObjects 是一种痛苦。您可以获取 JSON 响应的示例并将其粘贴到 json2csharp.com 之类的转换器中。它将为您生成一个类，然后您可以像这样使用它：

生成类：

public class MyClass
{
    public string SomeProperty { get; set; }
    public string AnotherProperty { get; set; }
}

用法：

if (message.IsSuccessStatusCode)
{
     var deserializedObject = JsonConvert.DeserializeObject<MyClass>(response.Content.ReadAsStringAsync().Result);
     Console.WriteLine(deserializedObject.SomeProperty);
}