小编Dav*_*ton的帖子

我有以下代码:

3.times do |n|
    "project#{n}" = FactoryGirl.create(:project, :title => "Project #{n}")
end

这显然不起作用...有谁知道如何制作循环,我可以使用'本地循环变量'来改变变量名称,以生成很多变量,如project1,project2,projekt3？

我正在使用异步任务来运行JSON下载器:(删节)

public class JSONDownloader extends AsyncTask<Object, Object, Object>{
    @Override
    protected Object doInBackground(Object... params) {
        if(JSONstate == false){
            try {
                final URL url = new URL([REDACTED]);

                final URLConnection urlConnection = url.openConnection();
                urlConnection.setRequestProperty("Content-Type", "application/json; charset=utf-8");
                urlConnection.connect();
                final InputStream inputStream = urlConnection.getInputStream();
                final StringBuilder sb = new StringBuilder();
                while (inputStream.available() > 0) {
                    sb.append((char) inputStream.read());
                }
                String result = sb.toString();

                JSONObject jsonOrg = new JSONObject(result);

                String ok = "ok";

                Response = jsonOrg.getString("response");
                System.out.println(Response);

                if(Response.equals(ok)){
                    Settingsresponse = true;

                    orgName = jsonOrg.getString("orgName");
                    System.out.println("orgName" + orgName); …

我下载了一个java程序,它包含两个文件夹src和分别包含源文件和类文件的类.现在,src和classes文件夹包含几个嵌套的子文件夹,其中最后一个子文件夹分别包含源文件和类文件.更确切地说,源文件和类文件的路径是src/edu/univ/.java和classes/edu/univ/.class.鉴于包含main函数的文件是Main.java,如何从命令行运行该程序.

我试过了:

 java  src/edu/univ/Main but I get Exception in thread "main" java.lang.NoClassDefFoundError: src/edu/univ/Main

我也尝试过:java src.edu.univ.Main但我遇到了类似的错误

请考虑以下代码:

    public Fingerprint(HashMap<String, Integer> measurements) {
        this();
        mMeasurements = measurements;
    }

   public Fingerprint(HashMap<String, Integer> measurements, String map) {
        this(measurements);
        mMap = map;
    }

    public Fingerprint(int id, String map, PointF location) {
        this();
        mLocation = location;
    }

    public Fingerprint(int id, String map, PointF location, HashMap<String, Integer> measurements) {
        this(id, map, location);
        mMeasurements = measurements;
    }

这样做的目的是什么(); 在这种背景下？因为我认为"this"指的是当前对象的字段.这是相同的定义吗？

试图采用数字的arraylist并打印出以下内容......

MOST POPULAR NUMBERS
The following numbers were picked 263 times: 41

LEAST POPULAR NUMBERS
The following numbers were picked 198 times: 20

AVERAGE
The Average was 228.545455 times.
The following numbers were picked 228 times:  5 22
The following numbers were picked 229 times:  2  7 12 40

我的代码......

import java.util.*;
import java.io.*;
import java.util.Arrays;
import java.util.Collections;
public class Hmwk {

    public static void main(String[] args) throws FileNotFoundException {
        Scanner input=new Scanner (new File ("input.txt"));
        int counter = 0; …

在Java中是否可以创建HashMap使用引用相等而不是equals()方法？

我的团队决定使用Struts 2.x,但我很困惑,因为Struts 1.x动作是单例,像servlet这样的动作是多线程的.Struts 2.x在每个请求上创建新实例,这会增加堆内存使用量.

使用Struts 2.x需要更多内存吗？

试图在程序之下运行时,它会在"消耗"之后被绞死.

1 : 2 : 3 : 4 : 5 : 6 : 7 : 8 : 9 : 10 : object realsed
object .....realsed
in consume..

public class MyThreadNew {

    public static void main(String[] args) {
        Object obj=new Object();
        boolean produce=true;
        Thread t1=new Thread(new P(obj,produce));
        Thread t2=new Thread(new C(obj,produce));

        t1.start();
        t2.start();
    }
}

class P implements Runnable{
    public Object obj;
    public static boolean produce;

    public P(Object obj, boolean produce){
        this.obj=obj;
        P.produce=produce;
    }

    public void produce(){
        synchronized (obj) {
            for(int …

复制受给定长度限制的列表模式的最佳方法是什么？

例如:

给定一个模式[1,2,3]和长度7,结果将是[1, 2, 3, 1, 2, 3, 1]

谁能告诉我这里做错了什么？我想要做的是循环遍历列表并为奇数和偶数分配背景颜色.

提前致谢.

var box = document.querySelectorAll('.links-wrapper li');


for(var i = 0; i < box.length; i++){
    if(box[i] % 2 === 0){
    	box[i].style.backgroundColor = 'red';
    } else {
    	box[i].style.backgroundColor = 'blue';
    }
}