我在一本书中看到了以下陈述:
任何基于写入复制
Iterator或ListIterator(例如添加,设置或删除)调用的变异方法都会抛出UnsupportedOperationException.
但是,当我运行以下代码时,它工作正常,并没有抛出UnsupportedOperationException.
List<Integer> list = new CopyOnWriteArrayList<>(Arrays.asList(4, 3, 52));
System.out.println("Before " + list);
for (Integer item : list) {
System.out.println(item + " ");
list.remove(item);
}
System.out.println("After " + list);
上面的代码给出了以下结果:
Before [4, 3, 52]
4
3
52
After []
为什么我在list使用该remove方法修改给定时没有得到异常？
美国/东部时区夏令时于 11 月 1 日凌晨 2 点结束。结果，凌晨 2 点变成了凌晨 1 点。
我无法理解下面给出的代码中的以下内容：
为什么第 4 行显示时间 10:00，而不是 09:00（添加 24 小时）？
LocalDateTime ld = LocalDateTime.of(2015, Month.OCTOBER, 31, 10, 0);
ZonedDateTime date = ZonedDateTime.of(ld, ZoneId.of("US/Eastern"));
System.out.println(date); //line 1 - 2015-10-31T10:00-04:00[US/Eastern]
date = date.plus(Duration.ofDays(1));
System.out.println(date); //line 2 - 2015-11-01T09:00-05:00[US/Eastern]
date = ZonedDateTime.of(ld, ZoneId.of("US/Eastern"));
System.out.println(date); //line 3 - 2015-10-31T10:00-04:00[US/Eastern]
date = date.plus(Period.ofDays(1));
System.out.println(date); //line 4 - 2015-11-01T10:00-05:00[US/Eastern]
有人可以帮我吗？
我无法${}在我的.jsp页面上运行表达式.
displayAllCustomers.jsp
<%@ taglib uri="http://java.sun.com/jsp/jstl/core" prefix="c" %>
<html>
<body>
<h3>Our Entire Customer Database</h3>
<ul>
<c:forEach items="${allCustomers}" var="customer">
<li>${customer.name}</li>
</c:forEach>
</ul>
</body>
</html>
调度员servlet.xml中
<?xml version="1.0" encoding="UTF-8"?>
<beans xmlns="http://www.springframework.org/schema/beans"
xmlns:xsi="http://www.w3.org/2001/XMLSchema-instance"
xmlns:p="http://www.springframework.org/schema/p"
xmlns:aop="http://www.springframework.org/schema/aop"
xmlns:tx="http://www.springframework.org/schema/tx"
xsi:schemaLocation="http://www.springframework.org/schema/beans
http://www.springframework.org/schema/beans/spring-beans-3.0.xsd
http://www.springframework.org/schema/tx
http://www.springframework.org/schema/tx/spring-tx-3.0.xsd
http://www.springframework.org/schema/aop
http://www.springframework.org/schema/aop/spring-aop-3.0.xsd">
<import resource="applicationContext.xml"/>
<bean class="org.springframework.web.servlet.view.InternalResourceViewResolver">
<property name="prefix" value="/WEB-INF/views/" />
<property name="suffix" value=".jsp" />
</bean>
<bean name="/displayAllCustomers" class="mypackage.DisplayAllCustomersController">
<property name="customerManagementService" ref="customerManagementService" />
</bean>
</beans>
DisplayAllCustomersController.java
public class DisplayAllCustomersController {
private CustomerManagementService customerManagementService;
public void setCustomerManagementService(CustomerManagementService customerManagementService) {
this.customerManagementService = customerManagementService; …以下是解决依赖关系时spring spring 3的错误:
Multiple annotations found at this line:
- Missing artifact org.springframework.security:org.springframework.security.web:jar:3.0.3.RELEASE
- ArtifactTransferException: Failure to transfer
org.springframework.security:org.springframework.security.web:jar:3.0.3.RELEASE from http://
repository.springsource.com/maven/bundles/release was cached in the local repository, resolution will not be
reattempted until the update interval of com.springsource.repository.bundles.release has elapsed or updates are
forced. Original error: Could not transfer artifact org.springframework.security:org.springframework.security.web:jar:
3.0.3.RELEASE from/to com.springsource.repository.bundles.release (http://repository.springsource.com/maven/
bundles/release): connection timed out to http://repository.springsource.com/maven/bundles/release/org/
springframework/security/org.springframework.security.web/3.0.3.RELEASE/
org.springframework.security.web-3.0.3.RELEASE.jar
以下是我的pom文件中定义的spring security 3依赖项的方法:
<!-- Spring Security dependencies -->
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>org.springframework.security.acls</artifactId>
<version>${spring.security.version}</version>
</dependency>
<dependency>
<groupId>org.springframework.security</groupId>
<artifactId>org.springframework.security.config</artifactId> …以下代码适用于Hibernate 4.3,但是当我使用Hibernate 5.0尝试相同的代码时,会导致以下错误:
Exception in thread "main" org.hibernate.MappingException: Unknown entity: entity.Message
at org.hibernate.internal.SessionFactoryImpl.getEntityPersister(SessionFactoryImpl.java:776)
at org.hibernate.internal.SessionImpl.getEntityPersister(SessionImpl.java:1451)
at org.hibernate.event.internal.AbstractSaveEventListener.saveWithGeneratedId(AbstractSaveEventListener.java:100)
at org.hibernate.event.internal.DefaultSaveOrUpdateEventListener.saveWithGeneratedOrRequestedId(DefaultSaveOrUpdateEventListener.java:192)
at org.hibernate.event.internal.DefaultSaveEventListener.saveWithGeneratedOrRequestedId(DefaultSaveEventListener.java:38)
at org.hibernate.event.internal.DefaultSaveOrUpdateEventListener.entityIsTransient(DefaultSaveOrUpdateEventListener.java:177)
at org.hibernate.event.internal.DefaultSaveEventListener.performSaveOrUpdate(DefaultSaveEventListener.java:32)
at org.hibernate.event.internal.DefaultSaveOrUpdateEventListener.onSaveOrUpdate(DefaultSaveOrUpdateEventListener.java:73)
at org.hibernate.internal.SessionImpl.fireSave(SessionImpl.java:678)
at org.hibernate.internal.SessionImpl.save(SessionImpl.java:670)
at org.hibernate.internal.SessionImpl.save(SessionImpl.java:665)
at client.Main.main(Main.java:14)
有人可以帮我理解为什么会这样吗？
以下是用于测试示例的代码:
Main.java
package client;
import org.hibernate.Session;
import util.HibernateUtil;
import entity.Message;
public class Main {
public static void main(String[] args) {
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
Message message = new Message( "Hello Hibernate 5" );
session.save(message);
session.getTransaction().commit();
session.close();
}
}
HibernateUtil.java
package util;
import org.hibernate.SessionFactory; …即使在用户点击浏览器后退按钮后,是否可以继续显示同一页面？
它就像一个多步骤表单,在提交时会重定向到结果页面,一旦表单提交,我不希望用户能够使用浏览器后退按钮导航回任何以前的页面.因此,我希望在单击后退按钮时向用户显示相同的结果页面.只需单击后退按钮的URL可以从历史记录更改为以前的URL,但我不希望用户返回到任何以前的页面.
如果可能的话,有人可以建议我如何实现这一点.
编辑:说有3个步骤的一些过程,然后生成的页面一样../step1,../step2,../step3,../resulting page.用户可以使用后退和前进按钮来播放这些步骤,但是一旦提交了步骤3并且用户在,../resulting page我希望用户仅在保持结果页面的同时显示先前显示的页面的URL.
我正在使用Hibernate 4,我想快速测试@OrderColumn注释但是我在尝试这样做时遇到以下错误:
INFO [main] (SchemaExport.java:343) - HHH000227: Running hbm2ddl schema export
DEBUG [main] (SchemaExport.java:353) - Import file not found: /import.sql
DEBUG [main] (SqlStatementLogger.java:104) -
alter table Order
drop
foreign key FK48E972E19D16493
Hibernate:
alter table Order
drop
foreign key FK48E972E19D16493
ERROR [main] (SchemaExport.java:425) - HHH000389: Unsuccessful: alter table Order drop foreign key FK48E972E19D16493
ERROR [main] (SchemaExport.java:426) - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax …我在这里有这个plunkr ,它显示了一个可编辑的表格.
以下是表格的HTML代码:
<body ng-controller="MainCtrl">
<table style="width:100%">
<tr>
<th>Name</th>
<th>Is enabled?</th>
<th>Points</th>
</tr>
<tr ng-repeat="fooObject in fooObjects | orderBy:'points'">
<td><input ng-model="fooObject.name" ng-disabled="fooState!='EDIT'"/></td>
<td><input ng-model="fooObject.isEnabled" ng-disabled="fooState!='EDIT'"/></td>
<td><input ng-model="fooObject.points" ng-disabled="fooState!='EDIT'"/></td>
<td>
<a href="#" ng-click="handleEdit(fooObject, 'EDIT', $index)">Edit</a>
<a href="#" ng-click="handleEditCancel(fooObject, 'VIEW', $index)">Cancel</a>
</td>
</tr>
</table>
</body>
我希望Cancel行中的链接显示前一个状态,fooObject就好像该行从未被触及一样.
以下是AngularJS控制器中的代码,它似乎只要我"orderBy:'points'"在ng-repeat表达式中没有,但是不起作用:
app.controller('MainCtrl', function($scope) {
$scope.fooObjects = [
{"name": "mariofoo", "points": 65, "isEnabled": true},
{"name": "supermanfoo", "points": 47, "isEnabled": false},
{"name": "monsterfoo", "points": 85, "isEnabled": true}
]; …我无法理解为什么BinaryOperator<Integer>可以A在下面的代码中放置在的位置，而不是BiFunction<Integer, Integer>？
A foo = (a, b) -> { return a * a + b * b; };
int bar = foo.apply(2, 3);
System.out.println(bar);
有人可以帮助我理解它。
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