i want to login to application from java code. I have found the following code example: String requestURL = "http://applicationURL/login.jsp"; String data = "email=temail@mail.com&password=123password&login=Login"; public static String sendPostRequest(String data, String requestURL) { String result=""; try { // Send the request URL url =
我想从java代码登录到应用程序。 我找到了下面的代码示例： String requestURL = "http://applicationURL/login.jsp"; String data = "email=temail@mail.com&password=123password&login=Login"; public static String sendPostRequest(String data, String requestURL) { String result=""; try { // Send the request URL url = new URL(requestURL);
In the following Spring 3.1 action, I've to do some stuff and add attribute to a POST request, and then redirect it to external URL through POST (I can't use GET). @RequestMapping(value = "/selectCUAA", method = RequestMethod.POST) public ModelAndView selectCUAA(@RequestParam(value="userID", required=true) String cuaa, ModelMap model) { //query & other... model.addAttribute(P
在下面的Spring 3.1操作中，我必须做一些事情并为POST请求添加属性，然后通过POST将其重定向到外部URL（我无法使用GET）。 @RequestMapping(value = "/selectCUAA", method = RequestMethod.POST) public ModelAndView selectCUAA(@RequestParam(value="userID", required=true) String cuaa, ModelMap model) { //query & other... model.addAttribute(PARAM_NAME_USER, cuaa); model.addAttribute(... , ...);
This question already has an answer here: Sending HTTP POST Request In Java 9 answers Convert Curl to Java equivalent 1 answer
这个问题在这里已经有了答案： 发送HTTP POST请求在Java 9中的答案 将Curl转换为Java相当于1个答案
This question already has an answer here: Sending HTTP POST Request In Java 9 answers you can use OkHttp https://github.com/square/okhttp public static final MediaType JSON = MediaType.parse("application/json; charset=utf-8"); OkHttpClient client = new OkHttpClient(); String post(String url, String json) throws IOException { RequestBody body = RequestBody.create(JSON, json); Requ
这个问题在这里已经有了答案： 发送HTTP POST请求在Java 9中的答案 你可以使用OkHttp https://github.com/square/okhttp public static final MediaType JSON = MediaType.parse("application/json; charset=utf-8"); OkHttpClient client = new OkHttpClient(); String post(String url, String json) throws IOException { RequestBody body = RequestBody.create(JSON, json); Request request = new Request
This question already has an answer here: Sending HTTP POST Request In Java 9 answers Make sure the "/Air" part is included in your url. Currently it's not.
这个问题在这里已经有了答案： 发送HTTP POST请求在Java 9中的答案 确保您的网址中包含“/ Air”部分。 目前不是。
I'm creating a game in Java for fun and I'm trying to decide how to organize my classes for the GUI. So far, all the classes with only the swing components and layout (no logic) are in a package called "ui". I now need to add listeners (ie ActionListener) to components (ie button). The listeners need to communicate with the Game class. Currently I have: Game.java - creates
我正在用Java创建一个游戏以获得乐趣，我正在试图决定如何组织我的GUI类。 到目前为止，只有swing组件和布局（无逻辑）的所有类都位于名为“ui”的包中。 我现在需要添加监听器（即ActionListener）组件（即按钮）。 听众需要与Game类进行通信。 目前我有：Game.java - 创建框架添加面板 import javax.swing.*; import ui.*; public class Game { private JFrame frame; Main main; Rules rules; Game()
I'm interested in hearing how you prefer to automate Javascript minification for your Java web apps. Here are a few aspects I'm particularly interested in: How does it integrate? Is it part of your build tool, a servlet filter, a standalone program post-processing the WAR file, or something else? Is it easy to enable and disable ? It's very unfunny to try and debug a minified s
我有兴趣听到你更喜欢为Java Web应用程序自动化Javascript缩小。 以下是我特别感兴趣的几个方面： 它如何整合？ 它是你的构建工具，servlet过滤器，后处理WAR文件的独立程序还是其他内容的一部分？ 启用和禁用是否容易 ？ 尝试和调试一个缩小的脚本是非常没有用的，但它对于开发人员能够测试缩小不会破坏任何东西也是有用的。 它是透明地工作，还是它有任何副作用（除了缩小的内在因素），我在日常工作中必须考虑这些副
I am new to REST API and I want to make a REST API call which returns a JSON object http://smlookup.dev/sec/products?search={"ABC.CP":"123A5"} - Runs fine in a browser and gives a JSON object how do i get '?search={"ABC.CP":"12345"}' this expression to work as it filter the records Code i am using is String serverUrl="http://smlookup.dev/
我是REST API的新手，我想创建一个返回JSON对象的REST API调用 http://smlookup.dev/sec/products?search={"ABC.CP":"123A5“} - 在浏览器中运行良好，并提供JSON对象 我如何得到'？search = {“ABC.CP”：“12345”}'这个表达式在过滤记录时起作用 我正在使用的代码是 String serverUrl="http://smlookup.dev/sec/products?search="; String search=URLEncoder.encode("={"ABC.CP":"12345"}")
I'm converting my data through REST protocol with JAX RS. I have a lot of working functionality already but I can't seem to convert the following thing: Function im calling public List<Tweet> search(String input) { ArrayList<Tweet> foundTweets = new ArrayList<Tweet>(); for(int i = 1; i <= this.tweets.size(); i++){ if(this.find(new Long(i)).getConte
我使用JAX RS通过REST协议转换我的数据。 我已经有很多工作功能，但我似乎无法转换以下内容： 函数即时调用 public List<Tweet> search(String input) { ArrayList<Tweet> foundTweets = new ArrayList<Tweet>(); for(int i = 1; i <= this.tweets.size(); i++){ if(this.find(new Long(i)).getContent().contains(input)){ foundTweets.add(this.tweets.get(i));
From the tutorial I had the impression that this should work (simplified example): public class Foo { private String bar; public String getBar() { return bar; } public void setBar(String bar) { this.bar = bar; } public static class Qux { private String foobar; public String getFoobar() { return foobar; } public v
从教程我有一个印象，这应该工作（简化的例子）： public class Foo { private String bar; public String getBar() { return bar; } public void setBar(String bar) { this.bar = bar; } public static class Qux { private String foobar; public String getFoobar() { return foobar; } public void setFoobar(String foobar) {