java - 在 Principle 上存储多个值

Question

我正在使用原理来存储当前用户 ID，

存储

   val principalExtractor: (MutableMap<String, Any?>) -> Any? = { params ->
        val session = params.get("session") as MutableMap<String, Any?>
        val user = session.get("user") as MutableMap<String, Any?>
        user.get("email")
    }

使用

@GetMapping("{userId}/activationWidgetUrl")
    fun getActivationWidgetUrl(principal: Principal, authentication: Authentication, @PathVariable userId: String): ActivationWidgetResponse? {

   let id = principal.name
}

问题:我面临的情况是，我需要 session 中的更多详细信息，例如:-companyId-。

我想在原理上存储一个 User 对象，而不仅仅是一个字符串，但principal.name 默认返回字符串。

问题:原则上如何存储对象(子类？)，或者有更好的方法吗？

Answer 1

通过使用身份验证解决了此问题。

principalExtractor

val principalExtractor: (MutableMap<String, Any?>) -> Any? = { params ->
        val session = params.get("session") as MutableMap<String, Any?>
        val user = session.get("user") as MutableMap<String, Any?>
        val mapper = ObjectMapper()
        mapper.configure(DeserializationFeature.FAIL_ON_UNKNOWN_PROPERTIES, false)
        val userDetails = mapper.convertValue(user, UserDetails::class.java)
        userDetails
    }

用法

@GetMapping("{userId}/activationWidgetUrl")
    fun getActivationWidgetUrl(authentication: Authentication, @PathVariable userId: String): ActivationWidgetResponse? {

       (authentication.principal as UserDetails).let { userDetails ->
            let id = userDetails.id
        }

}