使用RESTSHARP将XML进行反序列化
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12-12-2019 - |
题
我正在尝试为WP7项目尝试reakarp。在使用reakarp遇到某些XML时遇到麻烦。对象为null。这是一些相关的XML:
<?xml version="1.0" encoding="utf-8"?>
<api_response xmlns:xsd="http://www.w3.org/2001/XMLSchema" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance">
<response_data>
<employee_information>
<employee>
<employee_sf_name>David</employee_sf_name>
<employee_first_name>Dave</employee_first_name>
<employee_last_name>Jones</employee_last_name>
</employee>
</employee_information>
</response_data>
</api_response>
.
和这是我的要求:
public static void executeRequest(Action<string> callback, string method)
{
var client = new RestClient();
var request = new RestRequest(Method.POST);
client.BaseUrl = App.url + method;
request.AddParameter("secret_key", Application.secret_key);
request.AddParameter("email", Application.email);
request.AddParameter("password", Application.password);
client.ExecuteAsync<Employee>(request, response =>
{
callback(response.Content); //prints the response as output
Debug.WriteLine("firstname " + response.Data.employee_first_name);
});
}
.
以及员工对象:
public class Employee
{
public Employee() { }
public int employee_id { get; set; }
public String employee_first_name { get; set; }
public String employee_last_name { get; set; }
}
.
由于响应返回很好,我尝试在单独的函数中反序列化,但没有成功:
public static void parse(string data)
{
Debug.WriteLine(data);
XmlDeserializer xml = new XmlDeserializer();
Employee employee = new Employee();
employee = xml.Deserialize<Employee>(new RestResponse() { Content = data });
Debug.WriteLine("last name " + employee.employee_last_name);
Debug.WriteLine("firstname " + employee.employee_first_name);
}
.
提前感谢如果有人可以在问题上阐明一些灯光。
解决方案
首先,关闭标签需要是闭合标签。修复后,我设置了一个封闭类:
public class employee_information
{
public Employee employee { get; set; }
}
.
然后保留您的原始员工类:
public class Employee
{
public Employee() { }
public int employee_id { get; set; }
public String employee_first_name { get; set; }
public String employee_last_name { get; set; }
}
.
然后,将其反序列化:
var empInfo = xml.Deserialize<employee_information>((new RestResponse() {Content = data}));
. 其他提示
添加request.RootElement = "employee";
应该适用于现有代码。如果您不想从树下开始,则需要创建匹配整个层次结构的类。
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