How should I carefully construct url addresses to bypass forensics in this spring project?
public class SpringSecurityConfig
extends WebSecurityConfigurerAdapter {
@Bean
public HttpFirewall httpFirewall() {
return new CustomHttpFirewall();
}
protected void configure(HttpSecurity httpSecurity) throws Exception {
((ExpressionUrlAuthorizationConfigurer.AuthorizedUrl)httpSecurity.authorizeRequests().antMatchers(new String[]{"/admin/**"})).authenticated();
}
}
Above is the statement for authentication for this spring project.
And the project have the following statement in the firewall file.
public class CustomHttpFirewall
extends DefaultHttpFirewall {
protected void configure(StrictHttpFirewall firewalledRequest) {
firewalledRequest.setAllowUrlEncodedSlash(true);
firewalledRequest.setAllowUrlEncodedDoubleSlash(true);
firewalledRequest.setAllowUrlEncodedPeriod(true);
}
}
This project doesn't use JWT to authenticate a user.
The version number of spring-security in this project is 5.6.3, while which of springframework is 2.7.0
I've tried multiple ways of double-writing the slash and encoding the url multiple times, but all of them failed.
I wander whether there is even a way to bypass this forensic statement.