我一整天都在盯着这一点无济于事,我没有想法.IP-STS处理登录,然后将其传递给RP-STS,RP-STS用声明填充令牌,然后到网站上.这工作正常.我在IP-STS上有一些WCF功能,例如更改密码/重置密码,需要访问它们.阅读后,我应该能够将已经分配的令牌发送到WCF,以确保用户通过身份验证.从我可以看到它发送令牌但不正确,也不是我想要的方式,因为它似乎需要一个用户名.
理想情况下,我希望它获取分配给用户的令牌,而不必重新请求或重新创建任何内容,并且绝对不需要用户名/密码.
我目前得到的错误是: 从另一方收到了一个不安全或不正确安全的故障.请参阅内部FaultException以获取故障代码和详细信息.
详细信息: 无法处理消息.这很可能是因为"http://schemas.xmlsoap.org/ws/2005/02/trust/RST/Issue"操作不正确,或者因为邮件包含无效或过期的安全上下文令牌,或者因为存在不匹配绑定之间.如果服务由于不活动而中止了通道,则安全上下文令牌将无效.要防止服务中止空闲会话,请过早增加服务端点绑定的接收超时.
这是我的网络配置中的相关位.
客户端web.config
<system.serviceModel>
<bindings>
<customBinding>
<binding name="UsernameBinding"> <security authenticationMode="UserNameForCertificate" requireSecurityContextCancellation ="false" requireSignatureConfirmation="false" messageProtectionOrder ="SignBeforeEncryptAndEncryptSignature" requireDerivedKeys="true">
</security>
<httpTransport/>
</binding>
</customBinding>
<wsFederationHttpBinding>
<binding name="HTTPEndpoint" closeTimeout="00:10:00" openTimeout="00:10:00"
receiveTimeout="00:10:00" sendTimeout="00:10:00" bypassProxyOnLocal="false"
transactionFlow="false" hostNameComparisonMode="StrongWildcard"
maxBufferPoolSize="524288" maxReceivedMessageSize="65536" messageEncoding="Text"
textEncoding="utf-8" useDefaultWebProxy="true">
<security mode="Message">
<message issuedTokenType="http://docs.oasis-open.org/wss/oasis-wss-saml-token-profile-1.1#SAMLV1.1" negotiateServiceCredential="false"> <claimTypeRequirements>
<add claimType="http://schemas.xmlsoap.org/ws/2005/05/identity/claims/name" isOptional="false" />
</claimTypeRequirements>
<issuer address="http://localhost:13422/MembershipService" bindingConfiguration="UsernameBinding" binding="customBinding">
<identity>
<certificateReference findValue="STSTestCert" x509FindType="FindBySubjectName" storeLocation="CurrentUser" storeName="TrustedPeople" />
</identity>
</issuer>
</message>
</security>
</binding>
</wsFederationHttpBinding>
</bindings>
<behaviors>
<!-- Credentials configuration -->
<endpointBehaviors>
<behavior name="ServiceBehavior"> …Run Code Online (Sandbox Code Playgroud) 我目前正在尝试测试使用TryUpdateModel()的插入方法.我正在伪装控制器上下文,这是必要的,虽然它的工作原理似乎并没有张贴我设置的模型.
这是我正在测试的方法:
[AcceptVerbs(HttpVerbs.Post)]
[GridAction]
public ActionResult _SaveAjaxEditing(int? id)
{
if (id == null)
{
Product product = new Product();
if (TryUpdateModel(product))
{
//The model is valid - insert the product.
productsRepository.Insert(product);// AddToProducts(product);
}
}
else
{
var recordToUpdate = productsRepository.Products.First(m => m.ProductID == id);
TryUpdateModel(recordToUpdate);
}
productsRepository.Save();
return View(new GridModel(productsRepository.Products.ToList()));
}
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这是我目前的测试:
[TestMethod]
public void HomeControllerInsert_ValidProduct_CallsInsertForProducts()
{
//Arrange
InitaliseRepository();
var httpContext = CustomMockHelpers.FakeHttpContext();
var context = new ControllerContext(new RequestContext(httpContext, new RouteData()), controller);
controller.ControllerContext = context;
//controller.ControllerContext = new ControllerContext();
var …Run Code Online (Sandbox Code Playgroud)