将IP地址添加到HttpRequestMessage

Question

如何为HttpRequestMessage添加IP地址？

我正在为Web.API应用程序编写单元测试,我有一个AuthorizeAttribute来检查调用者的IP地址 -

        string[] AllowedIPs = new string[] { "127.0.0.1",  "::1" }
        string sourceIP = "";

        var contextBase = actionContext.Request.Properties["MS_HttpContext"] as System.Web.HttpContextBase;

        if (contextBase != null)
        {
            sourceIP = contextBase.Request.UserHostAddress;
        }

        if (!string.IsNullOrEmpty(sourceIP))
        {
            return AllowedIPs.Any(a => a == sourceIP);
        }
        return false;

我按如下方式构建测试请求 -

    var request = CreateRequest("http://myserver/api/CustomerController, "application/json", HttpMethod.Get);            
    HttpResponseMessage response = client.SendAsync(request).Result;

..

    private HttpRequestMessage CreateRequest(string url, string mthv, HttpMethod method)
    {
        var request = new HttpRequestMessage();
        request.RequestUri = new Uri(url);
        request.Headers.Accept.Add(new MediaTypeWithQualityHeaderValue(mthv));
        request.Method = method;
        return request;
    } 

但actionContext.Request.Properties ["MS_HttpContext"]为空,我无法执行测试.

以下解决方案 - 看看我自己的答案.

Answer 1

感谢Davin的建议.

这是解决方案 -

private HttpRequestMessage CreateRequest(string url, string mthv, HttpMethod method)
{
    var request = new HttpRequestMessage();

    var baseRequest = new Mock<HttpRequestBase>(MockBehavior.Strict);
    var baseContext = new Mock<HttpContextBase>(MockBehavior.Strict);

    baseRequest.Setup(br => br.UserHostAddress).Returns("127.0.0.1");
    baseContext.Setup(bc => bc.Request).Returns(baseRequest.Object);


    request.RequestUri = new Uri(url);
    request.Headers.Accept.Add(new MediaTypeWithQualityHeaderValue(mthv));

    request.Properties.Add("MS_HttpContext", baseContext.Object);

    request.Method = method;
    return request;
} 

然后以下列方式使用该方法 -

 var request = CreateRequest("http://myserver/api/CustomerController, "application/json", HttpMethod.Get);            
 HttpResponseMessage response = client.SendAsync(request).Result;