老师，进到了这个里面怎么出来
我自己创建的名为dev的collection中，capped属性为true，在进行update操作时总是提示更新后的数据大小要与原来的数据大小一致。
老师视频中创建的collection中的capped属性是false吧？
一、如下代码
package cn.sxt.pool108; import java.util.ArrayList; import java.util.List; import java.util.concurrent.Callable; import java.util.concurrent.ExecutionException; import java.util.concurrent.ExecutorService; import java.util.concurrent.Executors; import java.util.concurrent.Future; public class Test02 { public static void main(String[] args) throws InterruptedException, ExecutionException { // 如何创建一个线程池 // 1.创建一个线程池，线程池中只有一个线程对象 //ExecutorService pool = Executors.newSingleThreadExecutor(); // 2.创建一个线程池，线程池中有线程的数量固定 ExecutorService pool = Executors.newFixedThreadPool(10); // 3.创建一个线程池，线程池中的线程的数量可以动态的改变 //ExecutorService pool = Executors.newCachedThreadPool(); //创建一个集合 List<Future> list = new ArrayList<Future>(); /**使用线程池执行大量的Callable任务*/ for(int i=0;i<20;i++) { //使用匿名内部类 //创建任务 Callable<Integer> task = new Callable<Integer>() { @Override public Integer call() throws Exception { Thread.sleep(2000); return (int)(Math.random()*10)+1; } };//任务结束 //将任务交给线程池 Future f = pool.submit(task); //每执行完一个任务就将它添加到集合当中去 list.add(f); //System.out.println(f.get()); } System.out.println("ok?"); //任务全部执行完毕之后再去遍历集合 for (Future future : list) { System.out.println(future.get()); } System.out.println("OKOK!"); //关闭线程池 pool.shutdown(); } }
老师，当我将任务交给线程池以后，打印出来f.get();这时候效率没有增加，一个个打印出来。当我放入集合中的时候，也是每次一个个放进来，我并没有觉得效率增加了啊。为什么有System.out.println(f.get());的时候后面的线程要等System.out.println(f.get());完成后再走，而有list.add(f)时，线程可以同步运行。这不应该是同级别的吗？
老师这样怎么处理，浏览器打开一片空白，
<!DOCTYPE html> <html> <head lang="en"> <meta charset="UTF-8"> <title></title> </head> <body> <form action="" method="get"> <input type="text" name="userName"/> <input id="in" type="submit"/> </form> <button id="but">获取发送信息</button> <script> document.getElementById("but").onclick=function(){ console.log(document.location.search); } </script> </body> </html>
老师，现在这个js代码是点击
<button id="but">获取发送信息</button>
这个id=‘but’的button从而在控制台看到文本框的信息，
老师为什么不能把下面的onclick功能放到id=’in‘的那个submit上呢，就是点击一次就可以把文本框的信息输入到控制台中
老师，聊天室页面实现了，但是却有报错，点击发送也没有效果，图2的点击发送后的
这是源码：
源码.zip
<!DOCTYPE html> <html lang="en"> <head> <meta charset="UTF-8"> <meta name="viewport" content="width=device-width, initial-scale=1.0"> <title>Document</title> </head> <body> <ul> <li></li> <li></li> <li></li> <li></li> </ul> <script> var arr=['首页','游戏','直播','视频']; var ul=document.querySelector('ul'); var li=document.querySelectorAll('li'); for(var i=0;i<arr.length;i++){ li[i].innerHTML=arr[i]; } ul.onclick=function(){ li.style.cssText='background-color:red'; } </script> </body> </html>
老师哪里写错了吗
老师我有个疑问为什么有时候我无论对节点做什么操作，控制台都会报异常说找不到该节点，但再运行一次又行了呢？
一、问题：判断一个字符串的对称性。
老师，为啥我输出的会是false，感觉逻辑没有问题，而且代码跟老师讲课一样，但为什么输出的不是true。
二、代码
package com.bjsxt; import java.util.Stack; public class StackTest { public static void main(String[] args) { System.out.println("-------判断字符串的对称性---------"); StackTest stackTest = new StackTest(); stackTest.symmetry(); } //判断字符串的对成型 public void symmetry(){ String str="...{.....[....(....)...]....}..(....)..[...]..."; Stack<String> stack = new Stack<>(); boolean flag = true;//假设是匹配的 for (int i=0;i<str.length();i++){ char c = str.charAt(i); //拆分字符，把字符拿出来 if (c=='{'){ stack.push("}"); } if (c=='['){ stack.push("]"); } if (c=='('){ stack.push(")"); } if (c == '）'|| c == ']'|| c == '}'){ //修正处理 if (stack.empty()){ flag = false; break; } String x = stack.pop(); //修正处理 if (x.charAt(0) != c){ flag = false; break; } } } if (!stack.empty()){ flag = false; } System.out.println(flag); } }
三、运行效果图
if __name__ == '__main__': a1 = Account(100, "高淇") lock1 = Lock() draw1 = Drawing(80,a1) draw2 = Drawing(80,a1) draw1.start() draw2.start()
蓝莓派_project.zip
老师后台数据穿不来 前台也有问题
b=9.583 c=round(b) print(b) g=9.5 h=round(g) print(g)
老师为啥我这两串代码不是四舍五入的？该多少还是多少啊，咋回事？
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和视频里老师写的代码还是有点区别的，不过效果基本上都达到了
老师，我在pom里加jdbc的依赖了，还是爆红
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浏览器打开一片空白， 
哪里错了照着敲一直报错按着视频上一样桥的代码
和视频里老师写的代码还是有点区别的，不过效果基本上都达到了
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