使用spring webMVC和spring security登录ajax

Question

我一直使用Spring Security 3.0作为我们的网站登录机制,使用专用的登录网页.现在我需要登录网页,而不是我们站点中每个网页上的灯箱/弹出窗口,登录时我得到一个AJAX结果,无论它是否成功.使用Spring Security和Spring webmvc 3.0,最好的方法是什么？

Answer 1

在客户端,您可以通过ajax模拟正常表单提交到您的登录URL.例如,在jQuery中:

$.ajax({
    url: "${pageContext.request.contextPath}/j_spring_security_check",
    type: "POST",
    data: $("#loginFormName").serialize(),
    beforeSend: function (xhr) {
        xhr.setRequestHeader("X-Ajax-call", "true");
    },
    success: function(result) {
        if (result == "ok") {
            ...
        } else if (result == "error") {
            ...
        }
    }
});

在服务器端,您可以自定义AuthenticationSuccessHandler并AuthenticationFailureHandler返回值而不是重定向.因为您可能还需要一个普通的登录页面(尝试通过直接URL访问安全页面),您应该告诉正常调用的ajax调用,例如,使用标头:

public class AjaxAuthenticationSuccessHandler implements AuthenticationSuccessHandler {
    private AuthenticationSuccessHandler defaultHandler;

    public AjaxAuthenticationSuccessHandler() {

    }
    public AjaxAuthenticationSuccessHandler(AuthenticationSuccessHandler defaultHandler) {
        this.defaultHandler = defaultHandler;
    }

    public void onAuthenticationSuccess(HttpServletRequest request,
        HttpServletResponse response, Authentication auth)
        throws IOException, ServletException {
    if ("true".equals(request.getHeader("X-Ajax-call"))) {
        response.getWriter().print("ok");
        response.getWriter().flush();
    } else {
        defaultHandler.onAuthenticationSuccess(request, response, auth);
    }
}
}

Answer 2

我做了类似的事情(感谢axtavt):

public class AjaxAuthenticationSuccessHandler extends
    SimpleUrlAuthenticationSuccessHandler {

public void onAuthenticationSuccess(HttpServletRequest request,
        HttpServletResponse response, Authentication auth)
        throws IOException, ServletException {
    if ("XMLHttpRequest".equals(request.getHeader("X-Requested-With"))) {
        response.getWriter().print(
                "{success:true, targetUrl : \'"
                        + this.getTargetUrlParameter() + "\'}");
        response.getWriter().flush();
    } else {
        super.onAuthenticationSuccess(request, response, auth);
    }
}}

我选择为非Ajax请求扩展默认行为的简单成功处理程序.这是使XML工作的XML:

<http auto-config="false" use-expressions="true" entry-point-ref="authenticationProcessingFilterEntryPoint">
    <custom-filter position="FORM_LOGIN_FILTER" ref="authenticationFilter" />
...
...
</http>

<beans:bean id="authenticationProcessingFilterEntryPoint"
    class="org.springframework.security.web.authentication.LoginUrlAuthenticationEntryPoint">
    <beans:property name="loginFormUrl" value="/index.do" />
    <beans:property name="forceHttps" value="false" />
</beans:bean>

<beans:bean id="authenticationFilter" class=
    "org.springframework.security.web.authentication.UsernamePasswordAuthenticationFilter">
    <beans:property name="authenticationManager" ref="authenticationManager"/>
    <beans:property name="filterProcessesUrl" value="/j_spring_security_check"/>
    <beans:property name="sessionAuthenticationStrategy" ref="sas" />
    <beans:property name="authenticationFailureHandler" ref="failureHandler"/>
    <beans:property name="authenticationSuccessHandler" ref="successHandler"/>
</beans:bean>

<beans:bean id="successHandler" class="foo.AjaxAuthenticationSuccessHandler">
    <beans:property name="defaultTargetUrl" value="/login.html"/>
</beans:bean>

<beans:bean id="failureHandler" class="foo.AjaxAuthenticationFailureHandler" />