我在客户端有以下代码:
<script src="http://code.jquery.com/jquery-1.5.js"></script>
<script>
$(document).ready(function() {
$("a").click(function() {
//var orderId = $("#orderId").val();
$.post("test", { orderId : "John"},
function(data) {
alert("Data Loaded: " + data);
});
});
});
</script>
服务器端:
public void doPost(HttpServletRequest request, HttpServletResponse response) throws ServletException, IOException
{
PrintWriter writer = response.getWriter();
try{
String orderId = request.getAttribute("orderId").toString();
writer.write(orderId);
writer.close();
}
catch(Exception ex)
{
ex.getStackTrace();
}
}
我的
request.getAttribute("orderId")
是null并且我得到null引用exeption.我究竟做错了什么？
我有以下问题:我有一个查询,它返回35我的结果,我想保持在二级缓存:
public List<Product> getAllProducts() {
Session session = this.sessionfactory.getCurrentSession();
String queryString = "from com.ewave.upromotions.objects.Product product where product.active=:active";
Query query = session.createQuery(queryString);
query.setBoolean("active", true);
query.setCacheable(true);
query.setCacheRegion("productCache");
List<Product> products =query.list();
return products;
}
我的目标如下:
@Entity
@Table(name="products",schema="test11")
@Cacheable
@Cache( usage = CacheConcurrencyStrategy.NONSTRICT_READ_WRITE)
public class Product implements Serializable {
//all setters and getters ommited:
}
我的ehcache.xml文件位于/ src /目录中:
<ehcache>
<diskStore path="java.io.tmpdir"/>
<defaultCache maxElementsInMemory="10000"
eternal="false"
timeToIdleSeconds="120"
timeToLiveSeconds="120"
overflowToDisk="false"/>
<cache name="hibernate.test.org.hibernate.cache.UpdateTimestampsCache"
maxElementsInMemory="10000"
eternal="false"
timeToIdleSeconds="120"
timeToLiveSeconds="120"
overflowToDisk="true"/>
<cache name="hibernate.test.org.hibernate.cache.StandardQueryCache"
maxElementsInMemory="10000"
eternal="false"
timeToIdleSeconds="120"
timeToLiveSeconds="120"
overflowToDisk="true"/>
<cache name="com.vanilla.objects.Product"
maxElementsInMemory="300" …我只是在尝试乐观锁定.
我有以下课程:
@Entity
public class Student {
private Integer id;
private String firstName;
private String lastName;
private Integer version;
@Version
public Integer getVersion() {
return version;
}
//all other getters ommited.
}
现在我正在抓住其中一个学生并尝试同时更新其属性.
Thread t1 = new Thread(new MyRunnable(id));
Thread t2 = new Thread(new MyRunnable(id));
t1.start();
t2.start();
和MyRunnable内部:
public class MyRunnable implements Runnable {
private Integer id;
@Override
public void run() {
Session session = HibernateUtil.getSessionFactory().openSession();
session.beginTransaction();
Student student = (Student) session.load(Student.class, id);
student.setFirstName("xxxx");
session.save(student);
session.getTransaction().commit();
System.out.println("Done");
}
public MyRunnable(Integer …我在Controller中有一些执行@Async任务的方法
@Async
public Future<String> getResultFromServer(){
String result = ......
return new AsyncResult<String>(result);
}
方法执行时间最长为1分钟.我需要做的只是将结果返回给客户端,将使用AJAX/JQuery连接.
我不希望客户端每秒都请求我的服务器是否@Async执行方法.我只想保持连接打开,然后将结果"推送"到服务器.
@RequestMapping(value="/async.do", method=RequestMethod.POST)
public void getResult(HttpServletResponse res){
String result = null;
PrintWriter out = res.getWriter();
Future<String> future = getResultFromServer();
try {
if (future.isDone())
result = future.get();
out.println(result);
out.close();
} catch (InterruptedException e) {
e.printStackTrace();
} catch (ExecutionException e) {
e.printStackTrace();
}
}
我明白这与Comit模型非常接近,但我对彗星一般不熟悉.
我的问题是如何使用JavaScript/JQuery在客户端保持连接打开？
我的@RequestMapping(value="/async.do", method=RequestMethod.POST)方法会将结果推送给客户吗？
我在Hector和Cassandra教程中看到过DynamicCompositeType.
任何人都可以详细说明它们之间的区别
create column family Composite with comparator ='DynamicCompositeType
(t=>TimeUUIDType,s=>UTF8Type)'
and default_validation_class=UTF8Type and key_validation_class=UTF8Type;
和
create column family Composite
with comparator = 'CompositeType(TimeUUIDType,UTF8Type)'
and key_validation_class = 'UTF8Type'
and default_validation_class = 'UTF8Type'
我没有在Cassandra文档中找到它
这是我之前的问题的延续我如何在Akka接收方法上绕过类型擦除
我有10种类型的事件,它从我需要处理的事件扩展而来.
我想在单独的特征中为每个事件实现业务逻辑,因为混合所有10个事件处理函数将产生数百(如果不是数千)行代码.
我不想为每个事件创建不同的Actor类型.例如:
class Event1Actor extend Actor{
def receive ={
case Event1(e) => //event1 Business Logic
}
}
class Event2Actor extend Actor{
def receive ={
case Event2(e) => //event2 Business Logic
}
}
和相同的Event3Actor,Event4Actor等....
这样的代码对我来说似乎很难看,因为我需要在每个Actor中实现业务逻辑.
实现10个不同的特征和10个不同的Actor类似乎也是糟糕的设计.
我正在寻找基于设计模式的某种通用解决方案,例如策略模式.
我有以下JSF 2.1页面
<h:selectOneRadio value="#{userBean.newUser}">
<f:selectItem itemValue="0" itemLabel="new User" />
<f:selectItem itemValue="1" itemLabel="existing User" />
</h:selectOneRadio>
<br />
<h:inputText value="#{userBean.customerId}" id="customerId" />
<h:message for="customerid" />
<br />
<h:inputText value="#{userBean.firstName}" id="firstName" />
<h:message for="fisrtName" />
<br />
<h:inputText value="#{userBean.lastName}" id="lastName" />
<h:message for="lastName" />
<br />
<h:commandButton value="Submit" action="#{userBean.login}" />
这是我的豆子:
public class UserBean {
private String customerId;
private String newUser= "0";
private String firstName;
private String lastName;
// Getters and seeters ommited.
}
我需要通过以下方式验证此表单:
如果选择"新用户"单选按钮,则应验证所有表单输入.如果选择"现有用户",我只需要验证客户ID.
我尝试了Hibernate验证,我还通过实现javax.faces.validator.Validator接口尝试了自定义验证器.
我能以某种方式实现这样的功能吗？
我正在尝试编写一个非常简单的程序,它将模仿简单的DeadLock,其中线程A等待由线程B锁定的资源A,线程B等待由线程A锁定的资源B.
这是我的代码:
//it will be my Shared resource
public class Account {
private float amount;
public void debit(double amount){
this.amount-=amount;
}
public void credit(double amount){
this.amount+=amount;
}
}
这是我的runnable,它在上面的资源上执行Operation:
public class BankTransaction implements Runnable {
Account fromAccount,toAccount;
float ammount;
public BankTransaction(Account fromAccount, Account toAccount,float ammount){
this.fromAccount = fromAccount;
this.toAccount = toAccount;
this.ammount = ammount;
}
private void transferMoney(){
synchronized(fromAccount){
synchronized(toAccount){
fromAccount.debit(ammount);
toAccount.credit(ammount);
try {
Thread.sleep(500);
} catch (InterruptedException e) {
e.printStackTrace();
}
System.out.println("Current Transaction Completed!!!");
}
}
}
@Override
public …我有以下Callable:
public class Worker implements Callable<Boolean>{
@Override
public Boolean call(){
boolean success=true;
//do Something
return success;
}
}
现在我正在执行它:
Worker worker - new Worker();
Future<Boolean> submit = executor.submit(worker);
我正在以一种hashMap的形式存储提交,以便在代码中的某个地方执行某些操作.
如何知道worker.call()函数中是否发生了异常？
请问submit.isCancelled(),如果发生假某种异常的,如果一切工作正常返回true？
我需要创建Job,它将:
这个cron表达式有效吗？
Date start = 12/20/2012;
Date endDate = 12/31/2017;
SimpleTrigger trigger = newTrigger()
.withIdentity("trigger3", "group1")
.startAt(startDate)
.withSchedule(cronSchedule("* * 17 0 0/2 *,SUN,MON").build())
.endAt(endDate)
.build;
请指教.