我正试图为我参与的一个项目开发一个RESTful new服务,这对我来说都是非常新的。
我正在尝试向uses服务器添加一个基本身份验证(我的想法是,每个使用该服务的应用程序都具有凭据)。不会有任何网络接口来访问它,至少在这一点上。我一直在学习spring (http://spring.io/guides/tutorials/rest/)的教程,试图改变一些东西,主要是因为我试图使用Spring的最新版本。
我被困在一个表面上似乎很简单的部分,即只允许用户使用有效的用户名和密码向系统插入一个新值,但是由于某种原因,在运行特定测试时,我一直会得到错误403。
这是我的配置:
@Configuration
@EnableWebSecurity
public class SecurityConfig extends WebSecurityConfigurerAdapter{
@Autowired
protected void configureGlobal(AuthenticationManagerBuilder auth) throws Exception {
auth.inMemoryAuthentication()
.withUser("user").password("password").roles("USER");
}
@Override
protected void configure(HttpSecurity http) throws Exception {
http.antMatcher("/aggregators/**").authorizeRequests()
.anyRequest().hasRole("USER")
.and().httpBasic();
}
}在测试类中,错误来自构建个人()构建JSON的类
private ResponseEntity<Individual> buildIndividual() {
HttpEntity<String> requestEntity = new HttpEntity<String>(
RestDataFixture.standardIndividualJSON2(),
getHeaders("user:password"));
RestTemplate template = new RestTemplate();
ResponseEntity<Individual> entity = template.postForEntity(
"http://localhost:8080/aggregators/individuals",
requestEntity, Individual.class);
return entity;
}
static HttpHeaders getHeaders(String auth) {
HttpHeaders headers = new HttpHeaders();
headers.setContentType(MediaType.APPLICATION_JSON);
headers.setAccept(Arrays.asList(MediaType.APPLICATION_JSON));
byte[] codedString = Base64.encode(auth.getBytes());
headers.add("Authorization", "Basic " + new String(codedString));
return headers;
}函数standardndividualJSON2()只返回一个JSON字符串,如:"{“键”:"12345“,”名称“:"StrackOverflow”}“。
我还有一种方法试图创建一个密码错误的个人,我得到的错误也是403 (如果我没有弄错的话,它应该是401 )。这就是为什么我认为配置可能是错误的,但我不明白为什么。
我的配置有问题吗?如果有人能帮我,那就太好了。
谢谢你的帮助!
编辑
刚决定添加初始化程序类,可能问题是:
public class WebAppInitializer implements WebApplicationInitializer{
private static Logger LOG = LoggerFactory.getLogger(WebAppInitializer.class);
@Override
public void onStartup(ServletContext servletContext)
throws ServletException {
WebApplicationContext rootContext = createRootContext(servletContext);
configureSpringMvc(servletContext, rootContext);
configureSpringSecurity(servletContext, rootContext);
}
private WebApplicationContext createRootContext(ServletContext servletContext) {
AnnotationConfigWebApplicationContext rootContext = new AnnotationConfigWebApplicationContext();
rootContext.register(CoreConfig.class, SecurityConfig.class);
rootContext.refresh();
servletContext.addListener(new ContextLoaderListener(rootContext));
servletContext.setInitParameter("defaultHtmlEscape", "true");
return rootContext;
}
private void configureSpringMvc(ServletContext servletContext, WebApplicationContext rootContext) {
AnnotationConfigWebApplicationContext mvcContext = new AnnotationConfigWebApplicationContext();
mvcContext.register(MVCConfig.class);
mvcContext.setParent(rootContext);
ServletRegistration.Dynamic appServlet = servletContext.addServlet(
"webservice", new DispatcherServlet(mvcContext));
appServlet.setLoadOnStartup(1);
Set<String> mappingConflicts = appServlet.addMapping("/");
if (!mappingConflicts.isEmpty()) {
for (String s : mappingConflicts) {
LOG.error("Mapping conflict: " + s);
}
throw new IllegalStateException(
"'webservice' cannot be mapped to '/'");
}
}
private void configureSpringSecurity(ServletContext servletContext, WebApplicationContext rootContext) {
FilterRegistration.Dynamic springSecurity = servletContext.addFilter("springSecurityFilterChain",
new DelegatingFilterProxy("springSecurityFilterChain", rootContext));
springSecurity.addMappingForUrlPatterns(null, true, "/*");
}
}编辑2
通过禁用CSRF解决:
@Override
protected void configure(HttpSecurity http) throws Exception {
http.csrf().disable().
antMatcher("/aggregators/**").authorizeRequests()
.anyRequest().hasRole("USER")
.and().httpBasic();
}发布于 2015-01-16 15:36:37
通过禁用CSRF解决:
@Override
protected void configure(HttpSecurity http) throws Exception {
http.csrf().disable().
antMatcher("/aggregators/**").authorizeRequests()
.anyRequest().hasRole("USER")
.and().httpBasic();
}https://stackoverflow.com/questions/24189105
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