via*_*tor 23 authentication ajax jquery spring-security session-timeout
我在我的应用程序中使用spring-security和jQuery.主页使用Ajax动态加载内容到选项卡.一切都很好,但有时我的选项卡中有登录页面,如果我输入凭据,我将被重定向到没有标签的内容页面.
所以我想处理这种情况.我知道有些人使用ajax身份验证,但我不确定它是否适合我,因为它对我来说看起来很复杂,我的应用程序不允许任何访问而无需登录.我想为所有ajax响应编写一个全局处理程序,window.location.reload()
如果我们需要进行身份验证,它将会执行.我认为在这种情况下,最好是获取401
错误而不是标准登录表单,因为它更容易处理.
所以,
1)是否可以为所有jQuery ajax请求编写全局错误处理程序?
2)我如何自定义spring-security的行为以便为ajax请求发送401错误,但是常规请求像往常一样显示标准登录页面?
3)你可能有更优雅的解决方案吗?请分享.
谢谢.
Mat*_*man 10
这是一种我认为非常简单的方法.这是我在这个网站上观察到的一系列方法.我写了一篇关于它的博客文章:http: //yoyar.com/blog/2012/06/dealing-with-the-spring-security-ajax-session-timeout-problem/
基本思想是使用如上所述的api url前缀(即/ api/secured)以及认证入口点.它很简单而且很有效.
这是身份验证入口点:
package com.yoyar.yaya.config;
import org.springframework.security.core.AuthenticationException;
import org.springframework.security.web.authentication.LoginUrlAuthenticationEntryPoint;
import javax.servlet.ServletException;
import javax.servlet.http.*;
import java.io.IOException;
public class AjaxAwareAuthenticationEntryPoint
extends LoginUrlAuthenticationEntryPoint {
public AjaxAwareAuthenticationEntryPoint(String loginUrl) {
super(loginUrl);
}
@Override
public void commence(
HttpServletRequest request,
HttpServletResponse response,
AuthenticationException authException)
throws IOException, ServletException {
boolean isAjax
= request.getRequestURI().startsWith("/api/secured");
if (isAjax) {
response.sendError(403, "Forbidden");
} else {
super.commence(request, response, authException);
}
}
}
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这是你的spring上下文xml中的内容:
<bean id="authenticationEntryPoint"
class="com.yoyar.yaya.config.AjaxAwareAuthenticationEntryPoint">
<constructor-arg name="loginUrl" value="/login"/>
</bean>
<security:http auto-config="true"
use-expressions="true"
entry-point-ref="authenticationEntryPoint">
<security:intercept-url pattern="/api/secured/**" access="hasRole('ROLE_USER')"/>
<security:intercept-url pattern="/login" access="permitAll"/>
<security:intercept-url pattern="/logout" access="permitAll"/>
<security:intercept-url pattern="/denied" access="hasRole('ROLE_USER')"/>
<security:intercept-url pattern="/" access="permitAll"/>
<security:form-login login-page="/login"
authentication-failure-url="/loginfailed"
default-target-url="/login/success"/>
<security:access-denied-handler error-page="/denied"/>
<security:logout invalidate-session="true"
logout-success-url="/logout/success"
logout-url="/logout"/>
</security:http>
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我使用了以下解决方案.
在spring security中定义了无效的session url
<security:session-management invalid-session-url="/invalidate.do"/>
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对于该页面添加了以下控制器
@Controller
public class InvalidateSession
{
/**
* This url gets invoked when spring security invalidates session (ie timeout).
* Specific content indicates ui layer that session has been invalidated and page should be redirected to logout.
*/
@RequestMapping(value = "invalidate.do", method = RequestMethod.GET)
@ResponseBody
public String invalidateSession() {
return "invalidSession";
}
}
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并且对于ajax使用ajaxSetup来处理所有ajax请求:
// Checks, if data indicates that session has been invalidated.
// If session is invalidated, page is redirected to logout
$.ajaxSetup({
complete: function(xhr, status) {
if (xhr.responseText == 'invalidSession') {
if ($("#colorbox").count > 0) {
$("#colorbox").destroy();
}
window.location = "logout";
}
}
});
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我通常是这样做的。每次 AJAX 调用时,请在使用之前检查结果。
$.ajax({ type: 'GET',
url: GetRootUrl() + '/services/dosomething.ashx',
success: function (data) {
if (HasErrors(data)) return;
// process data returned...
},
error: function (xmlHttpRequest, textStatus) {
ShowStatusFailed(xmlHttpRequest);
}
});
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然后这个HasErrors()
功能就像这样,并且可以在所有页面上共享。
function HasErrors(data) {
// check for redirect to login page
if (data.search(/login\.aspx/i) != -1) {
top.location.href = GetRootUrl() + '/login.aspx?lo=TimedOut';
return true;
}
// check for IIS error page
if (data.search(/Internal Server Error/) != -1) {
ShowStatusFailed('Server Error.');
return true;
}
// check for our custom error handling page
if (data.search(/Error.aspx/) != -1) {
ShowStatusFailed('An error occurred on the server. The Technical Support Team has been provided with the error details.');
return true;
}
return false;
}
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