Can anyone give me an example of truly stateless RESTful endpoints? a simple question, if server is completely stateless, how do we invalidate previous tokens? I consider saving state to DB as bad practice. lets say there are hundreds of requests per second, that would mean hundreds of queries to DB per second (if you save state to DB) and that's bad news. if you save state to server, you
任何人都可以给我一个真正的无状态RESTful端点的例子吗？ 一个简单的问题，如果服务器是完全无状态的，我们如何使以前的令牌无效？ 我认为将状态保存到DB是一种不好的做法。 可以说每秒有数百个请求，这意味着每秒对数据库的数百次查询（如果将状态保存到数据库），那就是坏消息。 如果您将状态保存到服务器，则在使用多个服务器和负载平衡器时会遇到会话传输问题。 当然，一个例子当然是不需要验证的端点，并且依赖于你的
Say I have a URL http://example.com/query?q= and I have a query entered by the user such as: random word £500 bank $ I want the result to be a properly encoded URL: http://example.com/query?q=random%20word%20%A3500%20bank%20%24 What's the best way to achieve this? I tried URLEncoder and creating URI/URL objects but none of them come out quite right. URLEncoder should be the way to
假设我有一个URL http://example.com/query?q= 并且我有一个用户输入的查询，例如： 随机词500英镑银行$ 我希望结果是一个正确编码的网址： http://example.com/query?q=random%20word%20%A3500%20bank%20%24 达到此目的的最佳方式是什么？ 我试过URLEncoder并创建URI / URL对象，但没有一个出来很正确。 URLEncoder应该是要走的路。 您只需要记住只编码单个查询字符串参数名称和/或值，而不是整个URL，当然不是查询
This question already has an answer here: PUT vs. POST in REST 32 answers HttpServlet follow the HTTP specs as well which in simpler terms says: POST : To create a resource PUT : To modify a resource and if resource does not exist then create it HttpServlet provides an implementation of POST and PUT in doPost and doPut method. According to JavaDocs... HTTP POST method allows the cli
这个问题在这里已经有了答案： 在REST 32个答案中PUT vs. POST HttpServlet也遵循HTTP规范 ，用简单的术语说： POST ：创建资源 PUT ：修改资源，如果资源不存在，则创建它 HttpServlet在doPost和doPut方法中提供了POST和PUT的实现。 根据JavaDocs ... HTTP POST方法允许客户端一次性向Web服务器发送无限长度的数据，并且在发布诸如信用卡号码等信息时非常有用。 PUT操作允许客户端在服务器上放置一个文件，类似
i have some trouble with deep copying. I have this java project, chess, and I need to use the clone() method, because I need to try new configurations without changing the board. Board scacchiera = new Board(); Initialization(scacchiera); Board clone = scacchiera.clone(); System.out.println(scacchiera.toString()); System.out.println(clone.toString()); I create an object, sc
我在深拷贝时遇到了一些麻烦。 我有这个java项目，国际象棋，我需要使用克隆（）方法，因为我需要尝试新的配置，而无需更改电路板。 Board scacchiera = new Board(); Initialization(scacchiera); Board clone = scacchiera.clone(); System.out.println(scacchiera.toString()); System.out.println(clone.toString()); 我创建了一个对象，scacchiera，然后我克隆它。 我认为我已经正确完成了一个深层
I was learning about object deep cloning , I have an employee class with a getInstance method which returns a singleton and I am cloning the returned object.Below is the class and test class. public class Employee implements Serializable , Cloneable { public static Employee employee; private String name; private int age; private Employee(){ } public Employee(Str
我正在学习对象深层克隆，我有一个带有getInstance方法的employee类，它返回一个单例，并且克隆返回的对象。下面是类和测试类。 public class Employee implements Serializable , Cloneable { public static Employee employee; private String name; private int age; private Employee(){ } public Employee(String name, int age) { super(); this.name = name;
I tried to clone an object and change member functions of new clonned object. If it is shallow copy and according to wiki page, shallow copy and original object point to same object then , it should change the variable value. CloneExample obj1=new CloneExample(); CloneExample obj2=(CloneExample) obj1.clone(); obj1.a=2; obj1.c='a'; System.out.println("obj1 real "+obj1.a+" "+o
我试图克隆一个对象并更改新克隆对象的成员函数。 如果是浅拷贝，并且根据wiki页面，浅拷贝和原始对象指向同一个对象，则应该更改变量值。 CloneExample obj1=new CloneExample(); CloneExample obj2=(CloneExample) obj1.clone(); obj1.a=2; obj1.c='a'; System.out.println("obj1 real "+obj1.a+" "+obj1.c); System.out.println("obj2 real "+obj2.a+" "+obj2.c); obj2.a=99; obj2.c='z';
I am trying to modify account objects but the changes do not seem to appear in the original list afterward. Perhaps someone can pinpoint an error. see code below: if(aBank.getAccount(number)!=null){ System.out.println("Account information is listed below"); System.out.println(aBank.getAccount(number).toString()); System.o
我正在尝试修改帐户对象，但这些更改似乎未出现在原始列表中。 也许有人可以指出一个错误。 见下面的代码： if(aBank.getAccount(number)!=null){ System.out.println("Account information is listed below"); System.out.println(aBank.getAccount(number).toString()); System.out.println("Modify first name y or n");
I have an object which has many bufferedimages in it, I want to create a new object copying all the bufferedimages into the new object, but these new images may be altered and i don't want the original object images to be altered by altering the new objects images. is that clear? Is this possible to do and can anyone suggest a good way to do it please? I have thought of getSubImage but r
我有一个有许多bufferedimages的对象，我想创建一个新的对象，将所有的bufferedimages复制到新的对象中，但是这些新的图像可能会被改变，我不希望原始对象图像通过改变新的物体图像。 明白了吗？ 这有可能做到，任何人都可以提出一个好的方法来做到这一点吗？ 我曾想过getSubImage，但是在某处读取子图像的任何更改都会重新选回父图像。 我只是想能够得到一个完全独立的副本或一个BufferedImage的克隆 像这样？ static Bu
Before this question is marked as duplicate, please read it. ;) There are already several questions about coverage tools and such, however this is a bit different than the usual ones (I hope). According to wikipedia there are several different kind of 'coverage' variations that affect several different aspects of the term 'coverage'. Here a little example: public class Dummy
在此问题被标记为重复之前，请阅读它。 ;）关于报道工具等问题已经有几个问题，但是这与通常的有点不同（我希望）。 根据维基百科，有几种不同类型的“覆盖率”变化会影响“覆盖率”这个术语的几个不同方面。 这里有一个小例子： public class Dummy { public int a = 0; public int b = 0; public int c = 0; public void doSomething() { a += 5; b += 5; c = b + 5; } } pu
I have been doing some self-study on Big-O. I understand how to give examples of the following notations to algorithms: O(N): for(int i = 0; i < n; i++) sum++; O(N^2): for(int i = 0; i < n; i++) for( int j = 0; j < n; j++) sum++; O(N^3): for(int i = 0; i < n; i++) for( int j = 0; j < n * n; j++) sum++; I have come across these notations which I
我一直在做一些关于Big-O的自学。 我明白如何给算法提供以下符号的例子： 上）： for(int i = 0; i < n; i++) sum++; O（N ^ 2）： for(int i = 0; i < n; i++) for( int j = 0; j < n; j++) sum++; O（N ^ 3）： for(int i = 0; i < n; i++) for( int j = 0; j < n * n; j++) sum++; 我遇到了这些我不太了解的符号。 我如何根据算法给出这些例子？ 也许我应该这样说：