이것은 오래된 게시물이지만 "봄 보안 ajax 로그인"의 결과 중 하나로 나타나므로 내 솔루션을 공유 할 것이라고 생각했습니다. 이는 Spring Security 표준을 따르며 설정이 매우 간단합니다. 보안 구성에 REST/Ajax와 나머지 앱 (일반 HTML 페이지)에 대해 하나씩 2 <http> 요소를 갖는 것이 트릭입니다. <http>이 나타나는 순서는 중요하며 더 구체적인 URL에서 <http> 안에있는 <url-intercept> 개의 요소와 같은 방식으로 이루어져야합니다.
1 단계 : 설치 두 별도 <http>의
<?xml version="1.0" encoding="UTF-8"?>
<beans:beans xmlns="http://www.springframework.org/schema/security"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:beans="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://www.springframework.org/schema/beans http://www.springframework.org/schema/beans/spring-beans-3.1.xsd
http://www.springframework.org/schema/security http://www.springframework.org/schema/security/spring-security-3.1.xsd">
<!-- a shared request cache is required for multiple http elements -->
<beans:bean id="requestCache" class="org.springframework.security.web.savedrequest.HttpSessionRequestCache" />
<!-- remove security from static resources to avoid going through the security filter chain -->
<http pattern="/resources/**" security="none" />
<!-- http config for REST services (AJAX interface)
=================================================== -->
<http auto-config="true" use-expressions="true" pattern="/rest/**">
<!-- login configuration
login-processing-url="/rest/security/login-processing" front-end AJAX requests for authentication POST to this URL
login-page="/rest/security/login-page" means "authentication is required"
authentication-failure-url="/rest/security/authentication-failure" means "authentication failed, bad credentials or other security exception"
default-target-url="/rest/security/default-target" front-end AJAX requests are redirected here after success authentication
-->
<form-login
login-processing-url="/rest/security/login-processing"
login-page="/rest/security/login-page"
authentication-failure-url="/rest/security/authentication-failure"
default-target-url="/rest/security/default-target"
always-use-default-target="true" />
<logout logout-url="/rest/security/logout-url" />
<!-- REST services can be secured here, will respond with JSON instead of HTML -->
<intercept-url pattern="/rest/calendar/**" access="hasRole('ROLE_USER')" />
<!-- other REST intercept-urls go here -->
<!-- end it with a catch all -->
<intercept-url pattern="/rest/**" access="isAuthenticated()" />
<!-- reference to the shared request cache -->
<request-cache ref="requestCache"/>
</http>
<!-- http config for regular HTML pages
=================================================== -->
<http auto-config="true" use-expressions="true">
<form-login
login-processing-url="/security/j_spring_security_check"
login-page="/login"
authentication-failure-url="/login?login_error=t" />
<logout logout-url="/security/j_spring_security_logout" />
<intercept-url pattern="/calendar/**" access="hasRole('ROLE_USER')" />
<!-- other intercept-urls go here -->
<!-- in my app's case, the HTML config ends with permitting all users and requiring HTTPS
it is always a good idea to send sensitive information like passwords over HTTPS -->
<intercept-url pattern="/**" access="permitAll" requires-channel="https" />
<!-- reference to the shared request cache -->
<request-cache ref="requestCache"/>
</http>
<!-- authentication manager and other configuration go below -->
</beans:beans>
2 단계 : REST 인증 제어기
import org.springframework.http.HttpHeaders;
import org.springframework.http.HttpStatus;
import org.springframework.http.ResponseEntity;
import org.springframework.security.core.Authentication;
import org.springframework.security.core.context.SecurityContextHolder;
import org.springframework.stereotype.Controller;
import org.springframework.web.bind.annotation.RequestMapping;
import org.springframework.web.bind.annotation.RequestMethod;
import flexjson.JSONSerializer;
@Controller
@RequestMapping(value = "/rest/security")
public class RestAuthenticationController {
public HttpHeaders getJsonHeaders() {
HttpHeaders headers = new HttpHeaders();
headers.add("Content-Type", "application/json");
return headers;
}
@RequestMapping(value="/login-page", method = RequestMethod.GET)
public ResponseEntity<String> apiLoginPage() {
return new ResponseEntity<String>(getJsonHeaders(), HttpStatus.UNAUTHORIZED);
}
@RequestMapping(value="/authentication-failure", method = RequestMethod.GET)
public ResponseEntity<String> apiAuthenticationFailure() {
// return HttpStatus.OK to let your front-end know the request completed (no 401, it will cause you to go back to login again, loops, not good)
// include some message code to indicate unsuccessful login
return new ResponseEntity<String>("{\"success\" : false, \"message\" : \"authentication-failure\"}", getJsonHeaders(), HttpStatus.OK);
}
@RequestMapping(value="/default-target", method = RequestMethod.GET)
public ResponseEntity<String> apiDefaultTarget() {
Authentication authentication = SecurityContextHolder.getContext().getAuthentication();
// exclude/include whatever fields you need
String userJson = new JSONSerializer().exclude("*.class", "*.password").serialize(authentication);
return new ResponseEntity<String>(userJson, getJsonHeaders(), HttpStatus.OK);
}
}
3 단계 : AJAX 양식을 제출하고 응답을 처리 할 필요가 jQuery의 ajaxForm 라이브러리
<form action="/rest/security/login-processing" method="POST">
...
</form>
$('form').ajaxForm({
success: function(response, statusText, xhr, $form) {
console.log(response);
if(response == null || response.username == null) {
alert("authentication failure");
} else {
// response is JSON version of the Spring's Authentication
alert("authentication success");
}
},
error: function(response, statusText, error, $form) {
if(response != null && response.message == "authentication-failure") {
alert("authentication failure");
}
}
});
귀하의 솔루션은 좋은 것 같지만 곳 '에/휴식/보안/로그인 처리를 "정의 –
@MichaelDeKeyser 응답 컨트롤러 방법 -이 SpringSecurity에서 제공 UsernamePasswordAuthenticationFilter 매핑 것, 워드 프로세서 및 검색에서보세요 'UsernamePasswordAuthenticationFilter'및 'login-processing-url'에 대해 http://docs.spring.io/autorepo/docs/spring-security/3.2.0.RELEASE/reference/htmlsingle/ – SergeyB
Spring의 JavaConfig를 통해이 작업을 시도했습니다. 보안, 마침내 XML ftw! – Boon