小编ntv*_*t18的帖子

如何在 Scala 中模拟以下行为？即在满足蓄能器的某些特定条件时继续折叠。

def foldLeftWhile[B](z: B, p: B => Boolean)(op: (B, A) => B): B

例如

scala> val seq = Seq(1, 2, 3, 4)
seq: Seq[Int] = List(1, 2, 3, 4)
scala> seq.foldLeftWhile(0, _ < 3) { (acc, e) => acc + e }
res0: Int = 1
scala> seq.foldLeftWhile(0, _ < 7) { (acc, e) => acc + e }
res1: Int = 6

更新：

根据@Dima 的回答，我意识到我的意图有点副作用。所以我让它与 同步takeWhile，即如果谓词不匹配，则不会有任何进展。并添加更多示例以使其更清晰。（注意：这不适用于Iterators）

I have some docker containers provisioned by docker-compose with restart: always flag. But I accidentally delete docker-compose.yml file.

How do I delete those containers so that they will not restart automatically?

给定val as: Seq[Int] = ...

很多时候我需要对两个连续的元素应用操作，例如

顺便说一句我不喜欢

for (i <- 1 until as.size) {
  // do something with as(i) and as(i - 1)
}

或者通过另一个

as.tail.foldLeft((0, as.head)) { (acc, e) =>
  // do something with acc._2 and e 
  // and try to not forget returning (_, e) 
}

如何针对这种情况编写更好的代码？

我在 GCP 中有一个启用了 RBAC 的 Kubernetes 集群

有一个命名空间的分蘖和倍数服务

现在，我可以为特定服务帐户分配读者角色，因为它的全名

apiVersion: rbac.authorization.k8s.io/v1
kind: Role
metadata:
  annotations:
    rbac.authorization.kubernetes.io/autoupdate: "true"
  name: tiller-reader
  namespace: tiller
rules:
  - apiGroups: [""]
    resources: ["pods"]
    verbs:
    - "get"
    - "watch"
    - "list"
---
apiVersion: rbac.authorization.k8s.io/v1
kind: RoleBinding
metadata:
  name: tiller-reader-role-binding
  namespace: tiller
roleRef:
  apiGroup: rbac.authorization.k8s.io
  kind: Role
  name: tiller-reader
subjects:
  - apiGroup: rbac.authorization.k8s.io
    kind: User
    name: system:serviceaccount:my-namespace-1:my-namespace-1-service-account

该服务的命名空间和帐户动态创建。如何自动授予所有服务帐户访问Tiller 命名空间的权限，例如：获取 Pod？

为什么以下基本安全配置不适用于 inMemoryAuthentication() 子句？

@Configuration
@EnableWebSecurity
public class WebSecurityConfig extends WebSecurityConfigurerAdapter {

    @Override
    protected void configure(HttpSecurity http) throws Exception {
        http
            .csrf().disable()
            .httpBasic()
            .and()
            .authorizeRequests()
            .anyRequest().authenticated();
        super.configure(http);
    }

    @Override
    protected void configure(AuthenticationManagerBuilder auth) throws Exception {
        auth.inMemoryAuthentication()
            .withUser("username").password("password");
        super.configure(auth);
    }

}

应用初始化之后，仍然只有userSpring自己生成的default ，没有像username.

我有以下配置

@Configuration
@EnableWebFlux
public class WebfluxConfig {

    @Bean
    RouterFunction<?> routerFunction(UserResource userResource) {
        return route(GET("/user"), r -> ok()
                .body(userResource.findAll(), UserDto.class));
    }
}

当我启动应用程序时，它会进行预期的映射

2018-04-12 12:12:06.682 INFO 54617 --- [主要] oswrfssRouterFunctionMapping：映射/用户-> cnaicWebfluxConfig$$Lambda$764/909878836@14ba7f15

然而，当我尝试连接到它时，我得到了 404。并且路由器谓词没有被调用一次。

curl -v localhost:8080/user

*   Trying ::1...
* TCP_NODELAY set
* Connected to localhost (::1) port 8080 (#0)

> GET /user HTTP/1.1
> Host: localhost:8080
> User-Agent: curl/7.54.0
> Accept: */*
>

< HTTP/1.1 404
< Content-Type: application/json;charset=UTF-8
< Transfer-Encoding: chunked
< Date: Thu, 12 Apr 2018 09:15:52 GMT …