我正在使用WCF REST入门套件中提供的非常棒的HttpClient。我有以下方法来处理HelloTxt API:
public UserValidateResponse Validate()
{
HttpClient client = new HttpClient(baseUrl);
HttpMultipartMimeForm form = new HttpMultipartMimeForm();
form.Add("app_key", this.AppKey);
form.Add("user_key", this.UserKey);
HttpResponseMessage response = client.Post("user.validate", form.CreateHttpContent());
return response.Content.ReadAsXmlSerializable<UserValidateResponse>();
}
我有一个很好的通用GetRequest方法,如下所示:
public T GetRequest<T>(string query)
{
HttpClient client = new HttpClient(baseUrl);
client.DefaultHeaders.UserAgent.AddString(@"http://www.simply-watches.co.uk/");
HttpResponseMessage response = client.Get(query);
response.EnsureStatusIsSuccessful();
T data = default(T);
try
{
data = response.Content.ReadAsXmlSerializable<T>();
return data;
}
catch (Exception ex)
{
Console.Write(String.Format("{0}: {1}", ex.Message, ex.InnerException.Message));
}
return data;
}
其好处是您可以将T作为响应类型传递,如下随机示例所示:
public List<User> GetUsers(int deptid)
{
string query = String.Format("department.getUsers?api_key={0}&dept_id={1}", this.APIKey, deptId);
return GetRequest<List<User>>(query);
}
我现在想要使用相同的通用样式POST方法,而不是GET,我确信我可以使用HttpContentExtensions,但我无法弄清楚如何将请求转换为HttpMultipartMimeForm。这就是我到目前为止所做的:
public T PostRequest<K, T>(string query, K request)
{
HttpClient client = new HttpClient(baseUrl);
// the following line doesn't work! Any suggestions?
HttpContent content = HttpContentExtensions.CreateDataContract<K>(request, Encoding.UTF8, "application/x-www-form-urlencoded", typeof(HttpMultipartMimeForm));
HttpResponseMessage response = client.Post(query, content);
response.EnsureStatusIsSuccessful();
T data = default(T);
try
{
data = response.Content.ReadAsXmlSerializable<T>();
return data;
}
catch (Exception ex)
{
Console.Write(String.Format("{0}: {1}", ex.Message, ex.InnerException.Message));
}
return data;
}
它将被称为:
UserValidateResponse response = PostRequest<UserValidateRequest, UserValidateResponse>("user.validate", new UserValidateRequest(this.AppKey, this.UserKey));
这是针对此API:http://hellotxt.com/developers/documentation。任何建议都非常欢迎!我可以为每个POST定义一个不同的表单,但一般来说这样做会很好。
答案 0 :(得分:2)
我回答了我自己的问题。代码可以在我的.NET wrapper for the HelloTxt API - HelloTxt.NET中看到,根据我上面的评论,使用反射来计算请求对象属性,并使用值填充HttpMultipartMimeForm()
,同时检查Required
关于类属性的数据注释。
有问题的代码是:
/// <summary>
/// Generic post request.
/// </summary>
/// <typeparam name="K">Request Type</typeparam>
/// <typeparam name="T">Response Type</typeparam>
/// <param name="query">e.g. user.validate</param>
/// <param name="request">The Request</param>
/// <returns></returns>
public T PostRequest<K, T>(string query, K request)
{
using (var client = GetDefaultClient())
{
// build form data post
HttpMultipartMimeForm form = CreateMimeForm<K>(request);
// call method
using (HttpResponseMessage response = client.Post(query, form.CreateHttpContent()))
{
response.EnsureStatusIsSuccessful();
return response.Content.ReadAsXmlSerializable<T>();
}
}
}
/// <summary>
/// Builds a HttpMultipartMimeForm from a request object
/// </summary>
/// <typeparam name="T"></typeparam>
/// <param name="request"></param>
/// <returns></returns>
public HttpMultipartMimeForm CreateMimeForm<T>(T request)
{
HttpMultipartMimeForm form = new HttpMultipartMimeForm();
Type type = request.GetType();
PropertyInfo[] properties = type.GetProperties();
foreach (PropertyInfo property in properties)
{
foreach (Attribute attribute in property.GetCustomAttributes(true))
{
RequiredAttribute requiredAttribute = attribute as RequiredAttribute;
if (requiredAttribute != null)
{
if (!requiredAttribute.IsValid(property.GetValue(request, null)))
{
//Console.WriteLine("{0} [type = {1}] [value = {2}]", property.Name, property.PropertyType, property.GetValue(property, null));
throw new ValidationException(String.Format("{0} [type = {1}] requires a valid value", property.Name, property.PropertyType));
}
}
}
if (property.PropertyType == typeof(FileInfo))
{
FileInfo fi = (FileInfo)property.GetValue(request, null);
HttpFormFile file = new HttpFormFile();
file.Content = HttpContent.Create(fi, "application/octet-stream");
file.FileName = fi.Name;
file.Name = "image";
form.Files.Add(file);
}
else
{
form.Add(property.Name, String.Format("{0}", property.GetValue(request, null)));
}
}
return form;
}