其中一个(Ubuntu 16.04)谷歌云虚拟机的其中一个卷的磁盘利用率几乎一直是100% - 这是从系统中随机抽取的10秒样本:
iostat -x 10
Device: rrqm/s wrqm/s r/s w/s rkB/s wkB/s avgrq-sz avgqu-sz await r_await w_await svctm %util
sdd 0.60 17.20 5450.50 2468.00 148923.60 25490.00 44.05 11.81 1.49 1.13 2.29 0.13 99.60
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这是目前的2.5T持久SSD.
我的理解是,通过添加虚拟"主轴"然后在它们之间分配工作负载,我无法获得更好的性能.
这是一个数据库卷,所以我也不能真正使用易失性SSD.
我目前在XFS上有这些挂载选项:
type xfs (rw,noatime,nodiratime,attr2,nobarrier,inode64,noquota)
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有什么建议?
我有一个ListView,每次更改选择时,我想调用具有该名称的类.例如,如果该项称为"文本字符串",则应调用类TextString.我现在的代码给了我一个错误说The method insert(ArrayList<Element>) is undefined for the type Object... Eclipse给了我一个建议将对象强制转换为Element,但这没有做任何事情.Element类是一个超类,TextString将实现该类.
这是我到目前为止的代码:
elementList.itemsProperty().bind(listProperty);
listProperty.set(FXCollections.observableArrayList(elementListItems));
elementList.setOnMouseClicked(new EventHandler<MouseEvent>() {
public String selectedElement = "Text String";
@Override
public void handle(MouseEvent event) {
selectedElement = (String)elementList.getSelectionModel().getSelectedItem();
selectedElement = selectedElement.replace(" ", "");
Class<?> clazz;
try {
clazz = Class.forName("elements."+selectedElement);
Constructor<?> ctor = clazz.getConstructor();
Object object = ctor.newInstance();
Method meth = clazz.getClass().getMethod("insert", new Class<?>[] { Canvas.class, ArrayList.class, GraphicsContext.class });
meth.invoke(object, canvas, objects, gc);
} catch (ClassNotFoundException e) {
e.printStackTrace();
} catch (NoSuchMethodException e) {
e.printStackTrace(); …Run Code Online (Sandbox Code Playgroud) 我正在为我的应用程序使用 spring 数据休息。
当我在存储库中添加此方法时,出现以下错误并且应用程序无法启动:-
方法:-
@Modifying
@Transactional
@Query("from employee as ft where ft.company.id = ?1")
void deleteAllEmployeeCompany(
@Param("companyId") @RequestParam("companyId") int companyId);
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错误:-
org.springframework.context.ApplicationContextException: Failed to start bean 'documentationPluginsBootstrapper'; nested exception is java.lang.NullPointerException
at org.springframework.context.support.DefaultLifecycleProcessor.doStart(DefaultLifecycleProcessor.java:176) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.context.support.DefaultLifecycleProcessor.access$200(DefaultLifecycleProcessor.java:51) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.context.support.DefaultLifecycleProcessor$LifecycleGroup.start(DefaultLifecycleProcessor.java:346) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.context.support.DefaultLifecycleProcessor.startBeans(DefaultLifecycleProcessor.java:149) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.context.support.DefaultLifecycleProcessor.onRefresh(DefaultLifecycleProcessor.java:112) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.context.support.AbstractApplicationContext.finishRefresh(AbstractApplicationContext.java:874) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.boot.context.embedded.EmbeddedWebApplicationContext.finishRefresh(EmbeddedWebApplicationContext.java:144) ~[spring-boot-1.4.1.RELEASE.jar!/:1.4.1.RELEASE]
at org.springframework.context.support.AbstractApplicationContext.refresh(AbstractApplicationContext.java:544) ~[spring-context-4.3.3.RELEASE.jar!/:4.3.3.RELEASE]
at org.springframework.boot.context.embedded.EmbeddedWebApplicationContext.refresh(EmbeddedWebApplicationContext.java:122) ~[spring-boot-1.4.1.RELEASE.jar!/:1.4.1.RELEASE]
at org.springframework.boot.SpringApplication.refresh(SpringApplication.java:761) [spring-boot-1.4.1.RELEASE.jar!/:1.4.1.RELEASE]
at org.springframework.boot.SpringApplication.refreshContext(SpringApplication.java:371) [spring-boot-1.4.1.RELEASE.jar!/:1.4.1.RELEASE]
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如何解决这个问题?
更新:-
所有依赖项的列表:-
[INFO] com.test:test-service:jar:0.0.1-SNAPSHOT
[INFO] +- org.springframework.boot:spring-boot-starter-actuator:jar:1.4.1.RELEASE:compile
[INFO] | +- org.springframework.boot:spring-boot-starter:jar:1.4.1.RELEASE:compile
[INFO] | | …Run Code Online (Sandbox Code Playgroud) 我必须缓存以下public方法的结果:
@Cacheable(value = "tasks", key = "#user.username")
public Set<MyPojo> retrieveCurrentUserTailingTasks(UserInformation user) {
Set<MyPojo> resultSet;
try {
nodeInformationList = taskService.getTaskList(user);
} catch (Exception e) {
throw new ApiException("Error while retrieving tailing tasks", e);
}
return resultSet;
}
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我还在这里配置了缓存:
@Configuration
@EnableCaching(mode = AdviceMode.PROXY)
public class CacheConfig {
@Bean
public CacheManager cacheManager() {
final SimpleCacheManager cacheManager = new SimpleCacheManager();
cacheManager.setCaches(Arrays.asList(new ConcurrentMapCache("tasks"),new ConcurrentMapCache("templates")));
return cacheManager;
}
@Bean
public CacheResolver cacheResolver() {
final SimpleCacheResolver cacheResolver = new SimpleCacheResolver(cacheManager());
return cacheResolver;
}
}
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我断言如下: …
我正在使用spring AOP来建议我的服务方法,尤其是返回一个对象的方法,我想在建议处理过程中访问该对象.
我的配置工作正常,没有问题.
这是adviced方法的签名,方法根据方法参数中的数据返回一个新实例,因此参数不可用
@Traceable(ETraceableMessages.SAUVER_APPORTEUR)
public ElementNiveauUn save(ElementNiveauUn apporteur) throws ATPBusinessException {
String identifiant = instanceService.sauverInstance(null, apporteur);
List<String> extensions = new ArrayList<String>();
extensions.add(ELEMENTSCONTENUS);
extensions.add(TYPEELEMENT);
extensions.add(VERSIONING);
extensions.add(PARAMETRAGES);
extensions.add(PARAMETRAGES + "." + PARAMETRES);
return (ElementNiveauUn ) instanceService.lireInstanceParId(identifiant, extensions.toArray(new String[]{}));
}
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这就是我想要做的事情
@Around(value = "execution(elementNiveauUn fr.generali.nova.atp.service.metier.impl.*.*(..)) && @annotation(traceable) && args(element)", argNames = "element,traceable")
public void serviceLayerTraceAdviceBasedElementInstanceAfter2(final ProceedingJoinPoint pjp,
final ElementNiveauUn element, final Traceable traceable) throws SecurityException,
NoSuchMethodException, IllegalArgumentException, IllegalAccessException, InvocationTargetException {
// current user
String currentUserId = findCurrentUserId();
// wether user is found …Run Code Online (Sandbox Code Playgroud) 你好StackOverFlow(s)
我正在运行这个问题,因为现在超过2个小时这很简单
我正在尝试使用$ .ajax POST调用将JSON对象发送到Spring Controller
我正在使用AngularJS,但这一点很有用
这是服务器和客户端的代码以及弹簧配置
提前致谢
JQuery:
$scope.push = function() {
$.ajax({
type: "PUT",
url:"rest/todo/greeting/",
data : {id:"1",title:"ajax",description:"ajax"},
dataType: "json",
contentType : "application/json",
success : function(data) {
$log.info(data)
}
})
}
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弹簧控制器:
@Controller
@RequestMapping("/todo")
public class TodoController {
@RequestMapping(value = "/greeting", method = RequestMethod.PUT,consumes="application/json",produces="text/html")
public @ResponseBody String push(@RequestBody Todo todo) {
System.out.println(todo.getTitle());
return "test";
}
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}
弹簧配置:
<mvc:annotation-driven />
<context:component-scan base-package="org.lab.todo.controller" />
<bean id="defaultViews" class="org.springframework.web.servlet.view.json.MappingJackson2JsonView" />
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web.xml:
<?xml version="1.0" encoding="UTF-8"?>
<web-app version="2.5" xmlns="http://java.sun.com/xml/ns/javaee" xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance" xsi:schemaLocation="http://java.sun.com/xml/ns/javaee http://java.sun.com/xml/ns/javaee/web-app_2_5.xsd">
<listener> …Run Code Online (Sandbox Code Playgroud)