如何模拟 JWT.decode?

Question

我正在我的Spring引导应用程序中使用此库。

<dependency>
<groupId>com.auth0</groupId>
<artifactId>java-jwt</artifactId>
<version>3.9.0</version>
</dependency>

如何对DecodedJWT jwt = JWT.decode(accessToken);进行单元测试？

我可以传入一个实际的令牌，但这不是正确的方法。

我在我的Spring Boot应用程序中有这个JwtAuthenticationFilter。

@Component
public class JwtAuthenticationFilter extends OncePerRequestFilter {
@Value("${clientid}")
private String clientid;
@Autowired
private AuthenticationService authenticationService;
@Override
protected void doFilterInternal(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain) throws ServletException, IOException, InvalidRoleException {
getJwtFromRequest(request, response, filterChain);
}
private void getJwtFromRequest(HttpServletRequest request, HttpServletResponse response, FilterChain filterChain) throws ServletException, IOException {
String bearerToken = request.getHeader("Authorization");
if (!StringUtils.hasText(bearerToken) || !bearerToken.startsWith("Bearer ")) {
throw new AccessTokenMissingException("No access token found in request headers.");
}
try {
String accessToken = bearerToken.substring(7);
// this will also throw error when unable to reach auth server
ResponseEntity<String> result = authenticationService.getUserInfo(accessToken);
// Invalid access token
if (!result.getStatusCode().is2xxSuccessful()) {
throw new InvalidAccessTokenException("Invalid access token.");
}
DecodedJWT jwt = JWT.decode(accessToken);
String username = jwt.getClaim("preferred_username").asString();
Map<String, Object> resources = jwt.getClaim("resource_access").asMap();
Object roles = ((Map<String, Object>) resources.get(clientid)).get("roles");
List<String> rolesList = (ArrayList<String>)roles;
UserInfo user = new UserInfo();
user.setUsername(username);
user.setRole(rolesList);
// Step 3: Set username to security context
UsernamePasswordAuthenticationToken usernamePasswordAuthenticationToken = new UsernamePasswordAuthenticationToken(
user.getUsername(), null, AuthUtil.getAuthRole(user.getRole()));
SecurityContextHolder.getContext().setAuthentication(usernamePasswordAuthenticationToken);
} catch (HttpClientErrorException.Unauthorized | JWTDecodeException e) {
throw new InvalidAccessTokenException("Invalid access token.");
}
filterChain.doFilter(request, response);
}
}

Answer 1

您可以使用不同的策略/选项，所有这些都会起作用：

1. 根本不要嘲笑 JWT.decode，将其视为应在测试中运行的实用程序方法。它背后的直觉，如果你的班级使用Math.max(a,b)代码，或者一些日期时间操作，比如你会在测试中模拟它DateTime.of(...)呢？应该不会...尽管在这种情况下，您可能必须在测试中使用真正可解码的令牌

2. 使用 PowerMockito 来模拟静态调用(我真的不推荐这种方式，但如果您不想更改代码，它将完成这项工作(。

3. 执行重构，将解码功能提取到接口，并将其用作过滤器中的依赖项：

public interface JWTDecoder {
DecodedJWT decode(String token); // I assume its string for simplicity
}
@Component
public class StdJWTDecoder implements JWTDecoder {
public DecodedJWT decode(String token) {
return JWT.decode(tokent);
}
public class JwtAuthenticationFilter ... {
private final JWTDecoder jwtDecoder;
public JwtAuthenticationFilter(JWTDecoder jwtDecoder) {
this.jwtDecoder = jwtDecoder;
}
....
private void getJwtFromRequest(HttpServletRequest request, HttpServletResponse 
response, FilterChain filterChain)  {
...
// instead of:
DecodedJWT jwt = JWT.decode(accessToken);  
// use this:
DecodedJWT jwt = jwtDecoder.decode(accessToken);
...
}
}

使用这种方法，您可以轻松地用模拟来模拟JwtDecoder