我希望创建一个包含填充数据的HttpRequest实例,以便测试方法。
我需要将Microsoft.AspNetCore.Http.HttpRequest作为参数传递给函数。
我如何实例化它?
[Route("/summary/{id}")]
public IActionResult Account(int id)
{
var summary = RequestHelper.ParseRequest(Request);
}选项来自https://github.com/louthy/language-ext包
public static Option<SummaryRequest> ParseRequest(HttpRequest request)
{
if (request== null)
{
var query = request.Query;
var result = new SummaryRequest();
var locations = ExtractData(query, "location");
var categories = ExtractData(query, "categories[]");
var titles = ExtractData(query, "titles");
}
}
public static Option<SummaryRequest> ParseRequest(HttpRequest request)
{
if (request== null)
{
var query = request.Query;
var result = new SummaryRequest();
var locations = ExtractData(query, "location");
var categories = ExtractData(query, "categories[]");
var titles = ExtractData(query, "titles");
}
return new SummaryRequest();
}
public static Either<Exception, string[]> ExtractData(IEnumerable<KeyValuePair<String, StringValues>> query, string filter)
{
try
{
return query.First(x => x.Key.ToLower() == filter).Value.ToString().Split(',').ToArray();
}
catch (Exception ex)
{
return ex;
}
} 单元测试示例
[TestMethod]
[TestCategory(Test.RequestParser)]
public void ParseRequest_WithHttpRequest_ReturnResultOnSuccess()
{
// var request = this doesn't compile I need an instance of
// Microsoft.AspNetCore.Http.HttpRequest
var result = Helper.ParseRequest(request);
}发布于 2021-07-20 18:24:31
您可以使用Request property of DefaultHttpContext。这有点复杂,因为你必须单独指定很多属性(与WebRequest.Create相比),但下面的方法对我来说很有效:
var httpContext = new DefaultHttpContext();
httpContext.Request.Method = "POST";
httpContext.Request.Scheme = "http";
httpContext.Request.Host = new HostString("localhost");
httpContext.Request.ContentType = "application/json";
var stream = new MemoryStream();
var writer = new StreamWriter(stream);
await writer.WriteAsync("{}");
await writer.FlushAsync();
stream.Position = 0;
httpContext.Request.Body = stream;https://stackoverflow.com/questions/64634428
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