RestSharp 将复杂模型发送到 API，可以通过 FromForm 读取

Question

我在 ASP.NET Core 2.2 WebAPI 中有一个非常简单的方法：

[HttpPut]
public async Task<IActionResult> Put([FromForm] SimpleRequest request, [FromForm] string param1)
{
    if (request.ExtraData == null) return BadRequest("Please send data");
    await Task.CompletedTask;
    return Ok("Works");
}

我的模型是这样的：

public class SimpleRequest
{
    [Required]
    public string UserId { get; set; }
    public RequestType? RequestType { get; set; }
    public DateTime Timestamp { get; set; }
    public bool Test { get; set; }
    public ExtraData ExtraData { get; set; }
}
public enum RequestType
{
    Simple,
    Complex,
}
public class ExtraData
{
    public string FirstName { get; set; }
    public string LastName { get; set; }
    public ulong Id { get; set; }
    public DateTime RDate { get; set; }
    public uint Nr { get; set; }
}

当我向 Postman 提出请求时，我的模型绑定正确：

现在我尝试使用 RestSharp 创建相同的内容。当我使用下面的代码时它工作正常：

var client = new RestClient("https://localhost:44325/api/values");
var request = new RestRequest(Method.PUT);

request.AddParameter("UserId", "Sample", ParameterType.GetOrPost);
request.AddParameter("RequestType", "Simple", ParameterType.GetOrPost);
request.AddParameter("Timestamp", DateTime.Now, ParameterType.GetOrPost);
request.AddParameter("Test", true, ParameterType.GetOrPost);
request.AddParameter("ExtraData.FirstName", "John", ParameterType.GetOrPost);
request.AddParameter("ExtraData.LastName", "Smith", ParameterType.GetOrPost);
request.AddParameter("ExtraData.Id", 1234, ParameterType.GetOrPost);
request.AddParameter("ExtraData.Nr", 1, ParameterType.GetOrPost);
request.AddParameter("ExtraData.RDate", DateTime.Now.AddDays(-2), ParameterType.GetOrPost);

request.AddParameter("param1", "YES", ParameterType.GetOrPost);

var response = client.Execute(request);

但我想重用我的模型，因为它们在单独的 DLL 中共享。

我这样创建我的模型：

var model = new SimpleRequest
{
    UserId = "Sample",
    RequestType = RequestType.Simple,
    Timestamp = DateTime.Now,
    Test = true,
    ExtraData = new ExtraData
    {
        FirstName = "John",
        LastName = "Smith",
        Id = 1234,
        Nr = 1,
        RDate = DateTime.Now.AddDays(-2)
    }
};

但我无法通过请求传递此模型，以便它可以正确反序列化为我的对象。我试过使用 request.AddJsonBody 但失败了。

通过 RestSharp 传递上述模型的正确方法是什么？我应该如何序列化它（或添加为参数）才能使其正常工作。

Answer 1

对于FromForm，我们需要通过form-data或x-www-form-urlencoded发送请求。您可以实现自己的方式将模型转换为 x-www-form-urlencoded，例如：

要转换的扩展方法

public static class RestSharpExtension
{
        public static IDictionary<string, string> ToKeyValue(this object metaToken)
        {
        if (metaToken == null)
        {
                return null;
        }

        JToken token = metaToken as JToken;
        if (token == null)
        {
                return ToKeyValue(JObject.FromObject(metaToken));
        }

        if (token.HasValues)
        {
                var contentData = new Dictionary<string, string>();
                foreach (var child in token.Children().ToList())
                {
                var childContent = child.ToKeyValue();
                if (childContent != null)
                {
                        contentData = contentData.Concat(childContent)
                                                .ToDictionary(k => k.Key, v => v.Value);
                }
                }

                return contentData;
        }

        var jValue = token as JValue;
        if (jValue?.Value == null)
        {
                return null;
        }

        var value = jValue?.Type == JTokenType.Date ?
                        jValue?.ToString("o", CultureInfo.InvariantCulture) :
                        jValue?.ToString(CultureInfo.InvariantCulture);

        return new Dictionary<string, string> { { token.Path, value } };
        }
}

用例

public async Task<IActionResult> Index()
{
    var model = new SimpleRequest
                        {
                            UserId = "Sample",
                            RequestType = RequestType.Simple,
                            Timestamp = DateTime.Now,
                            Test = true,
                            ExtraData = new ExtraData
                            {
                                FirstName = "John",
                                LastName = "Smith",
                                Id = 1234,
                                Nr = 1,
                                RDate = DateTime.Now.AddDays(-2)
                            }
                        };
    var keyValueContent = model.ToKeyValue();
    keyValueContent.Add("param1", "YES");
    var formUrlEncodedContent = new FormUrlEncodedContent(keyValueContent);
    var urlEncodedString = await formUrlEncodedContent.ReadAsStringAsync();

    var client = new RestClient("http://localhost:51420/api/values/Put");
    var request = new RestRequest(Method.PUT);

    request.AddParameter("application/x-www-form-urlencoded", urlEncodedString, ParameterType.RequestBody);

    var response = client.Execute(request);
}

RestSharp 将复杂模型发送到 API，可以通过 FromForm 读取

ResrSharp send complex model to API that can be read via FromForm

.net

c#

restsharp

asp.net-core