我使用本教程在JavaFX 2中创建了自己的TreeView .但是我想隐藏我的根节点.
它现在看起来如何:
应该如何看待:
谢谢,Muki
我想匹配一些案例类.如果我不知道它们,我想匹配类必须扩展的指定特征.这看起来像
trait Event //root trait
trait Status extends Event //special trait
trait UIEvent extends Event //special trait
case class Results extends Event //concrete case class
case class Query extends Event //concrete case class
case class Running extends Status //concrete case class
case class Finished extends Status //concrete case class
case class Update extends UIEvent //concrete case class
我运行以下测试
val events = List(Results, Query, Running, Finished, Update)
events foreach {
case Results => println("Got a Results")
case Running => println("Got a …我有非常简单的汇编描述符
<assembly xmlns="http://maven.apache.org/plugins/maven-assembly-plugin/assembly/1.1.2" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xsi:schemaLocation="http://maven.apache.org/plugins/maven-assembly-plugin/assembly/1.1.2 http://maven.apache.org/xsd/assembly-1.1.2.xsd">
<id>dist</id>
<formats>
<format>jar</format>
</formats>
<!-- copied from jar-with-dependencies -->
<includeBaseDirectory>false</includeBaseDirectory>
<dependencySets>
<dependencySet>
<outputDirectory>/</outputDirectory>
<useProjectArtifact>true</useProjectArtifact>
<unpack>true</unpack>
<scope>runtime</scope>
</dependencySet>
</dependencySets>
<fileSets>
<fileSet>
<directory>${project.build.outputDirectory}</directory>
<outputDirectory>/</outputDirectory>
<useDefaultExcludes>true</useDefaultExcludes>
<excludes>
<exclude>**/dbAccess.*</exclude>
</excludes>
</fileSet>
</fileSets>
这主要是从Maven Descriptor Refs复制的.但是我组装的jar仍然包含dbAccess.*文件.我尝试了各种配置进行排除.
调试输出只是:
[DEBUG] The archive base directory is 'null'
[DEBUG] NOT reformatting any files in /media/truecrypt1/Development/workspace_java/baseanalysis/target/classes
[DEBUG] Adding file-set from directory: '/media/truecrypt1/Development/workspace_java/baseanalysis/target/classes'
assembly output directory is: ''
[DEBUG] Adding file-set in: /media/truecrypt1/Development/workspace_java/baseanalysis/target/classes to archive location:
如果我将它放在pom中的构建部分中,则正常的resouces过滤可以正常工作.
我想使用 github actions 来测试sbt-native-packager docker 集成，它从 Dockerfile 构建 docker 镜像。
问题是 github actions 似乎使用旧的或自定义的 docker 版本。3.0.8被检测为版本。查看失败的集成测试运行，其中指出
[1]检测到的Docker版本DockerVersion(3,0,8,None)与DockerPermissionStrategy.MultiStage不兼容
我想使用最新的 docker 版本 19.x，它允许我们测试所有功能。理想情况下，我能够为不同的测试场景设置不同的 docker 版本，但如果有的话那就太好了。
更新
docker version
Client:
Version: 3.0.8
API version: 1.40
Go version: go1.12.10
Git commit: 2355349d
Built: Wed Oct 23 17:47:59 2019
OS/Arch: linux/amd64
Experimental: false
Server:
Engine:
Version: 3.0.8
API version: 1.40 (minimum version 1.12)
Go version: go1.12.10
Git commit: adfac69
Built: Wed Oct 23 17:54:47 2019 …如何将期货与超时结合起来有一些很好的提示.但是我很好奇如何用Future sequence sequenceOfFutures做到这一点
我的第一种方法看起来像这样
import scala.concurrent._
import scala.concurrent.duration._
import scala.concurrent.ExecutionContext.Implicits._
object FutureSequenceScala extends App {
println("Creating futureList")
val timeout = 2 seconds
val futures = List(1000, 1500, 1200, 800, 2000) map { ms =>
val f = future {
Thread sleep ms
ms toString
}
Future firstCompletedOf Seq(f, fallback(timeout))
}
println("Creating waitinglist")
val waitingList = Future sequence futures
println("Created")
val results = Await result (waitingList, timeout * futures.size)
println(results)
def fallback(timeout: Duration) = future {
Thread …我想为值类派生Encoder实例.使用该semiauto机制,我无法派生嵌套类.
映像以下案例类结构
{
case class Recipient(email: Recipient.Email, name: Recipient.Name)
object Recipient {
case class Email(value: String) extends AnyVal
case class Name(value: String) extends AnyVal
}
}
在一个菊石壳(也添加收件人案例类)
load.ivy("io.circe" %% "circe-core" % "0.6.1")
load.ivy("io.circe" %% "circe-generic" % "0.6.1")
import io.circe._
import io.circe.generic.semiauto._
import io.circe.syntax._
现在Email为预期的结果导出解码器
Recipient.Email("ab@cd.com").asJson(deriveEncoder[Recipient.Email])
Json = {
"value" : "ab@cd.com"
}
派生Encoder[Recipient]不起作用
deriveDecoder[Recipient]
could not find Lazy implicit value of type
io.circe.generic.decoding.DerivedDecoder[$sess.cmd5.Recipient]
我想做的是派生一个Encoder[Recipient.Email]返回包装类型.如果我明确地派生出编解码器,这个小部分就可以工作了.
import shapeless.Unwrapped
implicit def encodeAnyVal[W <: AnyVal, U]( …我目前正在尝试使用Tree该类从数据库查询构建树结构.之后我想将它转换为json对象(使用playframework api).
Tree该类的一些示例或更多文档将是非常棒的.我无法得到我的头draw和drawTree方法,这可能做一个类似的事情.
val tree = ("Root", 100).node(
("Category1", 30).leaf,
("Category2", 20).node(
("Sub1", 15).leaf,
("Sub2", 3).leaf,
("Sub3", 2).leaf),
("Category3", 10).leaf,
("Category4", 30).node(
("Sub1", 20).leaf,
("Sub2", 5).leaf))
这应该会产生这样的json树
{
"name" : "Root",
"value" : 100,
"children" : [
{
"name" : "Category1",
"value": 30
},
{
"name": "Category2",
"value": 20,
"children" : [
{
"name" : "Sub1",
"value" : 15"
} ....
]
]
我们开始实现Source.queue[HttpRequest]文档中提到的模式:http://doc.akka.io/docs/akka-http/current/scala/http/client-side/host-level.html#examples
这是文档中的(简化)示例
val poolClientFlow = Http()
.cachedHostConnectionPool[Promise[HttpResponse]]("akka.io")
val queue =
Source.queue[(HttpRequest, Promise[HttpResponse])](
QueueSize, OverflowStrategy.dropNew
)
.via(poolClientFlow)
.toMat(Sink.foreach({
case ((Success(resp), p)) => p.success(resp)
case ((Failure(e), p)) => p.failure(e)
}))(Keep.left)
.run()
def queueRequest(request: HttpRequest): Future[HttpResponse] = {
val responsePromise = Promise[HttpResponse]()
queue.offer(request -> responsePromise).flatMap {
case QueueOfferResult.Enqueued => responsePromise.future
case QueueOfferResult.Dropped => Future.failed(new RuntimeException("Queue overflowed. Try again later."))
case QueueOfferResult.Failure(ex) => Future.failed(ex)
case QueueOfferResult.QueueClosed => Future.failed(new RuntimeException("Queue was closed (pool shut down) while running the request. Try …我正在使用Eclipse 3.7与Eclipse Gemini JPA/DBAccess.我实现了一个引用EntityManagerBuilderFactory的声明式ServiceComponent.bind方法如下所示:
public void bindEntityManagerFactoryBuilder(EntityManagerFactoryBuilder emfBuilder) {
emfFactoryBuilders.add(emfBuilder);
}
但是我需要ServiceReference来获取此服务的属性,例如:
ServiceReference ref = ...
String unitName = (String)ref.getProperty(EntityManagerFactoryBuilder.JPA_UNIT_NAME);
我是否必须在ServiceRegistry中搜索此ServiceReference,还是有更优雅的方法？
欢呼,Muki
我有一个单页webapp,大约300Kb.它与webpack捆绑在一起并使用React/Redux.
在页面加载之前,如何预先放置一个小的加载微调器？我可以成像