spring - 通过 URL 进行 Spring Security 身份验证

Question

我有一个 Spring MVC 应用程序，它使用 Spring Security 和基于表单的登录进行授权/身份验证。

现在我想添加一个特殊的 URL，其中包含一个无需其他信息即可访问的令牌，因为该令牌对用户来说是唯一的：

http://myserver.com/special/5f6be0c0-87d7-11e2-9e96-0800200c9a66/text.pdf

如何配置 Spring Security 以使用该令牌进行用户身份验证？

Answer 1

您可以提供自定义 PreAuthenticatedProcessingFilter 和 PreAuthenticatedAuthenticationProvider。有关详细信息，请参阅预身份验证方案一章。

Answer 2

您需要定义您的自定义预授权过滤器。

在http标签内的安全应用上下文中：

<custom-filter position="PRE_AUTH_FILTER" ref="preAuthTokenFilter" />

然后定义您的过滤器bean（及其适当的属性）：

<beans:bean class="com.yourcompany.PreAuthTokenFilter"
      id="preAuthTokenFilter">
    <beans:property name="authenticationDetailsSource" ref="authenticationDetailsSource" />
    <beans:property name="authenticationManager" ref="authenticationManager" />
    <beans:property name="authenticationEntryPoint" ref="authenticationEntryPoint"/>
</beans:bean>

创建从 GenericFilterBean 扩展的自定义过滤器

public class PreAuthTokenFilter extends GenericFilterBean {

private AuthenticationEntryPoint authenticationEntryPoint;
private AuthenticationManager authenticationManager;
private AuthenticationDetailsSource authenticationDetailsSource = new WebAuthenticationDetailsSource();

@Override
public void doFilter(ServletRequest req, ServletResponse resp,
                     FilterChain chain) throws IOException, ServletException {
    HttpServletRequest request = (HttpServletRequest) req;
    HttpServletResponse response = (HttpServletResponse) resp;

    String token = getTokenFromHeader(request);//your method

    if (StringUtils.isNotEmpty(token)) {
        /* get user entity from DB by token, retrieve its username and password*/

        if (isUserTokenValid(/* some args */)) {
            try {
                UsernamePasswordAuthenticationToken authRequest = new UsernamePasswordAuthenticationToken(username, password);
                authRequest.setDetails(this.authenticationDetailsSource.buildDetails(request));
                Authentication authResult = this.authenticationManager.authenticate(authRequest);
                SecurityContextHolder.getContext().setAuthentication(authResult);
            } catch (AuthenticationException e) {
            }
        }
    }

    chain.doFilter(request, response);
}

/*
other methods
*/

If you don't want or cannot retrieve a password, you need to create your own AbstractAuthenticationToken which will receive only username as param (principal) and use it instead of UsernamePasswordAuthenticationToken:

public class PreAuthToken extends AbstractAuthenticationToken {

    private final Object principal;

    public PreAuthToken(Object principal) {
        super(null);
        super.setAuthenticated(true);
        this.principal = principal;
    }

    @Override
    public Object getCredentials() {
        return "";
    }

    @Override
    public Object getPrincipal() {
        return principal;
    }
}

Answer 3

我遇到了这个问题，并使用 Spring Security RembereMe Service 基础架构的自定义实现解决了它。这是您需要做的。

• 定义自己的 Authentication 对象

  公共类 LinkAuthentication 扩展 AbstractAuthenticationToken { @Override public Object getCredentials() { return null; }

  @Override
  public Object getPrincipal()
  {
  
      return the prncipal that that is passed in via the constructor 
  }
  

  }

定义

public class LinkRememberMeService implements RememberMeServices, LogoutHandler
{    
    /**
     * It might appear that once this method is called and returns an authentication object, that authentication should be finished and the
     * request should proceed. However, spring security does not work that way.
     * 
     * Once this method returns a non null authentication object, spring security still wants to run it through its authentication provider
     * which, is totally brain dead on the part of Spring this, is why there is also a
     * LinkAuthenticationProvider
     * 
     */
    @Override
    public Authentication autoLogin(HttpServletRequest request, HttpServletResponse response)
    {
        String accessUrl = ServletUtils.getApplicationUrl(request, "/special/");
        String requestUrl = request.getRequestURL().toString();
        if (requestUrl.startsWith(accessUrl))
        {
            // take appart the url extract the token, find the user details object 
                    // and return it. 
            LinkAuthentication linkAuthentication = new LinkAuthentication(userDetailsInstance);
            return linkAuthentication;
        } else
        {
            return null;
        }
    }

    @Override
    public void loginFail(HttpServletRequest request, HttpServletResponse response)
    {
    }

    @Override
    public void loginSuccess(HttpServletRequest request, HttpServletResponse response, Authentication successfulAuthentication)
    {
    }

    @Override
    public void logout(HttpServletRequest request, HttpServletResponse response, Authentication authentication)
    {
    }
}


public class LinkAuthenticationProvider implements AuthenticationProvider
{

    @Override
    public Authentication authenticate(Authentication authentication) throws AuthenticationException
    {
        // Spring Security is totally brain dead and over engineered
        return authentication;
    }

    @Override
    public boolean supports(Class<?> authentication)
    {
        return LinkAuthentication.class.isAssignableFrom(authentication);
    }

}

破解其余的 spring security xml 以定义自定义身份验证提供程序和自定义记住我服务。

PS，如果您在 URL 中对 GUID 进行 base64 编码，它将缩短几个字符。您可以使用 Apache commons codec base64 二进制编码器/解码器来做更安全的 url 链接。

public static String toBase64Url(UUID uuid)
{
    return Base64.encodeBase64URLSafeString(toBytes(uuid));
}