SpringSecurity登陸受權出現This object has not been built問題解決

1、問題

在使用SpringSecurity作登陸受權時，出現了這樣的異常：java.lang.IllegalStateException: This object has not been built

/**
     * 登陸驗證
     *
     * @param username 用戶名
     * @param password 密碼
     * @param code     驗證碼
     * @param uuid     惟一標識
     * @return 結果
     */
    public String login(String username, String password, String code, String uuid) {
        String verifyKey = Constants.CAPTCHA_CODE_KEY + uuid;
        String captcha = redisCache.getCacheObject(verifyKey);
        redisCache.deleteObject(verifyKey);
        if (captcha == null) {
            AsyncManager.me().execute(AsyncFactory.recordLogininfor(username, Constants.LOGIN_FAIL, MessageUtils.message("user.jcaptcha.expire")));
            throw new CaptchaExpireException();
        }
        if (!code.equalsIgnoreCase(captcha)) {
            AsyncManager.me().execute(AsyncFactory.recordLogininfor(username, Constants.LOGIN_FAIL, MessageUtils.message("user.jcaptcha.error")));
            throw new CaptchaException();
        }
        // 用戶驗證
        Authentication authentication = null;
        try {
            // 該方法會去調用UserDetailsServiceImpl.loadUserByUsername
            /*
            authentication = authenticationManager
                    .authenticate(new UsernamePasswordAuthenticationToken(username, password));
             */

            UsernamePasswordAuthenticationToken authenObj = new UsernamePasswordAuthenticationToken(username, password);

            authentication = authenticationManager
                    .authenticate(authenObj);

        } catch (Exception e) {
            if (e instanceof BadCredentialsException) {
                AsyncManager.me().execute(AsyncFactory.recordLogininfor(username, Constants.LOGIN_FAIL, MessageUtils.message("user.password.not.match")));
                throw new UserPasswordNotMatchException();
            } else {
                AsyncManager.me().execute(AsyncFactory.recordLogininfor(username, Constants.LOGIN_FAIL, e.getMessage()));
                throw new CustomException(e.getMessage());
            }
        }
        AsyncManager.me().execute(AsyncFactory.recordLogininfor(username, Constants.LOGIN_SUCCESS, MessageUtils.message("user.login.success")));
        LoginUser loginUser = (LoginUser) authentication.getPrincipal();
        // 生成token
        return tokenService.createToken(loginUser);
    }

異常調試：

2、解決

目前該bug還在調試中...