我有一个 Web 方法,在查询字符串中有一个 token ,每个方法都使用此 token 来识别请求用户。
为此,我创建了一个名为 AuthenticationFilter 的 ResourceFilter,它从请求中获取 token 并从数据库中检索用户信息,如果用户有效,则允许执行该方法,否则将引发 UNAUTHORIZED 异常。
我的问题是:我如何访问在 AuthenticationFilter 类中检索到的用户信息
我的网络 API:
@GET
@Consumes(MediaType.TEXT_PLAIN)
@Produces({MediaType.APPLICATION_JSON})
@Path("/getUserOrders")
@ResourceFilters({AuthenticationFilter.class, AllowOroginFilter.class})
public String getUserOrders(@Context UriInfo uriInfo) {
//I need to access Usr object that retrived in AuthenticationFilter
/*User Usr = AuthenticationFilter.Usr;*/
String Result = getUserOrders(Usr);
return Result;
}
资源过滤器代码:
public class AuthenticationFilter implements ResourceFilter, ContainerRequestFilter, ContainerResponseFilter {
public User usr = null;
@Override
public ContainerRequest filter(ContainerRequest containerRequest) {
usr = AuthenticationUtiity.AuthenticateRequest(containerRequest);
if(usr == null){
Response.ResponseBuilder builder = null;
String response = "{\"Success\":false, \"Message\":\"Invalid username or password\"}";
builder = Response.status(Response.Status.UNAUTHORIZED).entity(response);
throw new WebApplicationException(builder.build());
}
return containerRequest;
}
}
我用谷歌搜索了这个问题,找到了一种方法可以使用 ThreadLocal 并将用户保存到 ThreadLocal 中并在该线程中的任何位置获取它。
https://veerasundar.com/blog/2010/11/java-thread-local-how-to-use-and-code-sample/
有没有更简单的方法来做到这一点?谢谢
最佳答案
您可以设置 SecurityContext
在ContainerRequest
上。然后在您的资源方法中,您可以使用 @Context SecurityContext
注入(inject)它。对于 SecurityContext
实现,您可以实现 Principal
并根据需要使其详细。当您在资源方法中获取 Principal
时,只需将其转换为您的类型即可。下面是一个过滤器示例:
@Provider
public class TestFilter implements ContainerRequestFilter {
@Override
public ContainerRequest filter(ContainerRequest request) {
SecurityContext oldSec = request.getSecurityContext();
final String username = "foobar";
final String email = "email@email.com";
final User user = new User(username, email);
request.setSecurityContext(new MySecurityContext(user, oldSec.isSecure()));
return request;
}
private static class MySecurityContext implements SecurityContext {
private final boolean isSecure;
private final User user;
public MySecurityContext(User user, boolean isSecure) {
this.isSecure = isSecure;
this.user = user;
}
@Override
public Principal getUserPrincipal() {
return this.user;
}
@Override
public boolean isUserInRole(String s) {
return false;
}
@Override
public boolean isSecure() {
return this.isSecure;
}
@Override
public String getAuthenticationScheme() {
return null;
}
}
public static class User implements Principal {
private final String email;
private final String username;
public User(String username, String email) {
this.username = username;
this.email = email;
}
@Override
public String getName() {
return null;
}
public String getEmail() {
return this.email;
}
}
}
然后在资源方法中
@Path("test")
public class TestResource {
@GET
public String get(@Context SecurityContext sc) {
TestFilter.User user = (TestFilter.User) sc.getUserPrincipal();
return user.getEmail();
}
}
更新
thank you for the answer but i dont want to add extra argument to my web method
然后将其作为字段注入(inject)
@Path("test")
public class TestResource {
@Context
private SecurityContext sc;
@GET
public String get() {
TestFilter.User user = (TestFilter.User) sc.getUserPrincipal();
return user.getEmail();
}
}
关于java - 如何访问 Web api 方法内的资源过滤器中生成的对象,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/51582999/