For jQuery Mobile I need markup like: <form action="..." method="get" data-ajax="false"> <!-- Fields --> </form> Since I work with Spring, I really like what <form:form> is doing for me, with all the convenient bindings, generating fields etc. How can I make <form:form> print the extra attribute? The <form:form> tag will allow arbitrary attributes. &
对于jQuery Mobile我需要像下面这样的标记： <form action="..." method="get" data-ajax="false"> <!-- Fields --> </form> 自从我和Spring一起工作以来，我真的很喜欢<form:form>为我做的事情，所有便利的绑定，生成字段等等。 我怎样才能让<form:form>打印额外的属性？ <form:form>标签将允许任意属性。 <form:form commandName="blah" data-ajax="false"> 将工作得很
Í have JNI function that needs to return a jbyte array. The array contains byte data of ac struct with data. But i get an error when i pass the data form the unsinged char array to the jbyte array. The C struct is defined as followed: // ObjectInfo struct definition struct ObjectInfo { int ObjectXCor; int ObjectYCor; int ObjectMass; }; // ObjectInfo struct definition struct
Í有JNI函数需要返回一个jbyte数组。 该数组包含带有数据的ac结构的字节数据。 但是当我将unsinged char数组的数据传递给jbyte数组时，我得到一个错误。 C结构定义如下： // ObjectInfo struct definition struct ObjectInfo { int ObjectXCor; int ObjectYCor; int ObjectMass; }; // ObjectInfo struct definition struct SensorDataStruct{ int PingData; int IRData; int ForceData; in
I have a weird error of double insert in the DB. I have the following classes: TestEntity - entity with a @PrePersist and a @PostPersist methods. Auditoria - audit entity Dataset - interface of DatasetBean DatasetBean - Stateless bean which implements Dataset DatasetFactory - instance an EJB of Dataset (lookup) I put the problem in a junit test (I'm using embedded Glassfish): @T
我在数据库中有一个奇怪的双重插入错误。 我有以下类： TestEntity - 使用@PrePersist和@PostPersist方法的实体。 Auditoria - 审计实体 数据集 - DatasetBean的接口 DatasetBean - 实现Dataset的无状态bean DatasetFactory - 实例数据集的EJB（查找） 我在junit测试中发现了这个问题（我正在使用嵌入式Glassfish）： @Test public void test() throws NamingException { Dataset<TestEntity> datas
public class Customer { private User user; private String name; public User getUser() { return user; } public void setUser(User user) { this.user = user; } public String getName() { return name; } public void setName(String name) { this.name = name; } } public class User { private String name; public String get
public class Customer { private User user; private String name; public User getUser() { return user; } public void setUser(User user) { this.user = user; } public String getName() { return name; } public void setName(String name) { this.name = name; } } public class User { private String name; public String get
I can serialize a List in my servlet on GAE, but I can't deserialize it. What am I doing wrong? This is my class Video in GAE, which is serialized: package legiontube; import java.util.Date; import javax.jdo.annotations.IdGeneratorStrategy; import javax.jdo.annotations.IdentityType; import javax.jdo.annotations.PersistenceCapable; import javax.jdo.annotations.Persistent; import javax.j
我可以在GAE上的我的servlet中序列化List，但是我无法反序列化它。 我究竟做错了什么？ 这是我的课堂视频GAE，这是连载： package legiontube; import java.util.Date; import javax.jdo.annotations.IdGeneratorStrategy; import javax.jdo.annotations.IdentityType; import javax.jdo.annotations.PersistenceCapable; import javax.jdo.annotations.Persistent; import javax.jdo.annotations.PrimaryKey; @Persisten
This question already has an answer here: Why use getters and setters/accessors? [closed] 38 answers There is also the point of view that most of the time, using setters still breaks encapsulation by allowing you to set values that are meaningless. As a very obvious example, if you have a score counter on the game that only ever goes up, instead of // Game private int score; public void se
这个问题在这里已经有了答案： 为什么使用getter和setter / accessors？ [已完成] 38个答案 还有一种观点认为，大多数时候，使用setter仍然允许你设置无意义的值来打破封装。 作为一个非常明显的例子，如果你的游戏中只有一个分数计数器，而不是 // Game private int score; public void setScore(int score) { this.score = score; } public int getScore() { return score; } // Usage game.setScore(game.getScore() +
I am reading "The Java Tutorial" (for the 2nd time). I just got through the section on Interfaces (again), but still do not understand how Java Interfaces simulate multiple inheritance. Is there a clearer explanation than what is in the book? Suppose you have 2 kinds of things in your domain : Trucks and Kitchens Trucks have a driveTo() method and Kitchens a cook() method. Now s
我正在阅读“The Java Tutorial”（第二次）。 我刚刚完成了关于Interfaces的一节，但仍不明白Java接口如何模拟多重继承。 有没有比书中的内容更清晰的解释？ 假设你的域中有两种东西：卡车和厨房 卡车有一个driveTo（）方法和Kitchens一个cook（）方法。 现在假设泡利决定从货车背面卖掉比萨饼。 他想要一个他可以开车去（）和做饭（）的东西。 在C ++中，他会使用多重继承来做到这一点。 在Java中被认为太危险了，
I have been told to develop my code according to the following code example: Class Dogs{ List<Dog> listOfDogs = new ArrayList<Dog>(); // Setters, getters... } Class Dog{ // Attributes and methods } The question is: why is it better to implement a Dogs class as a container for the ArrayList of Dog objects, instead of just referring to a simple ArrayList. Is this considered
我被告知根据以下代码示例开发我的代码： Class Dogs{ List<Dog> listOfDogs = new ArrayList<Dog>(); // Setters, getters... } Class Dog{ // Attributes and methods } 问题是：为什么将Dogs类作为Dog对象的ArrayList的容器来实现会更好，而不是仅仅引用一个简单的ArrayList。 这是否被认为是封装？ 基本上，创建Dogs类为您提供了额外的间接层。 这意味着你的所有其他代码将只引用（即使用） Do
Consider the following class: public class Cars extends Observable{ private ArrayList<String> carList = new ArrayList<String>(); public void addToCarList(String car){ // ... hasChanged(); notifyObservers(); } public void removeFromCarList(String car){ // ... hasChanged(); notifyObservers(); } public ArrayList
考虑以下课程： public class Cars extends Observable{ private ArrayList<String> carList = new ArrayList<String>(); public void addToCarList(String car){ // ... hasChanged(); notifyObservers(); } public void removeFromCarList(String car){ // ... hasChanged(); notifyObservers(); } public ArrayList<String>
This question already has an answer here: Is List<Dog> a subclass of List<Animal>? Why are Java generics not implicitly polymorphic? 15 answers This generic question in Java may look confusing to any one who is not very familiar with Generics as in first glance it looks like String is object so List<String> can be used where List<Object> is required but this is not t
这个问题在这里已经有了答案： List <Dog>是List <Animal>的子类吗？ 为什么Java泛型不是隐含多态的？ 15个答案 Java中的这个通用问题可能让任何不熟悉泛型的人感到困惑，因为乍看起来它看起来像String是对象，所以List<String>可以用在需要List<Object>的地方，但这不是真的。 这会导致编译错误。 因为List<Object> 可以存储任何东西，包括String ， Integer等，但List<String>只