如何反序列化json对象的特定键？

Question

输入JSON文件：

{
    "@version": "2.7.0",
    "@generated": "Wed, 30 May 2018 17:23:14",
    "site": {
        "@name": "http://google.com",
        "@host": "google.com",
        "@port": "80",
        "@ssl": "false",
        "alerts": [
            {

                "alert": "X-Content-Type-Options Header Missing",
                "name": "X-Content-Type-Options Header Missing",
                "riskcode": "1",
                "confidence": "2",
                "riskdesc": "Low (Medium)",
                "desc": "<p>The Anti-MIME-Sniffing header X-Content-Type-Options was not set to 'nosniff'. This allows older versions of Internet Explorer and Chrome to perform MIME-sniffing on the response body, potentially causing the response body to be interpreted and displayed as a content type other than the declared content type. Current (early 2014) and legacy versions of Firefox will use the declared content type (if one is set), rather than performing MIME-sniffing.</p>",
                "instances": [
                    {
                        "uri": "http://google.com",
                        "method": "GET",
                        "param": "X-Content-Type-Options"
                    }
                ],          
                "wascid": "15",
                "sourceid": "3"
            }

        ]
    }
}

预期输出:列出警报；

其中：

public class Alert
{
    public string alert;
    public string riskcode;
}

我想获取json对象的特定键，然后在alert对象中将其反序列化。

Answer 1

最简单的方法是只声明外部对象，并使用足够的键到达您关心的键：

public class Alert
{
    public string alert;
    public string riskcode;
}

public class SiteAlerts
{
    public Site site { get; set; }
}

public class Site
{
    public List<Alert> alerts { get; } = new List<Alert>();
}

然后，您可以简单地使用以下命令进行反序列化：

var siteAlerts = JsonConvert.DeserializeObject<SiteAlerts>(json);
var alerts = siteAlerts.site.alerts; // no error-checking here

Answer 2

我建议您使用Newtonsoft.Json库来简化json数据的反序列化。

如果你想要一个的部分反序列化，例如，只将alerts属性反序列化到你的类Alert中，而不需要创建所需的整个类结构。

您可以使用以下代码：

JObject jObject = JObject.Parse(json);
var alerts = jObject["site"]["alerts"].ToObject<Alert[]>();
foreach(var item in alerts)
{
    Console.WriteLine("alert: " + item.alert);  
    Console.WriteLine("riskcode: " + item.riskcode);
}

完整的演示可用here。

Answer 3

简短版本

var siteAlerts = JsonConvert.DeserializeObject<dynamic>(json).site.alerts.ToObject<Alert[]>();