Question

我编写了java类来计算给定网站的响应时间并将数据存储在文件中。所以我想调用正在计算的方法。它也有争论。当我在main方法中调用该方法时，它显示出一些错误。有人可以修复我的错误吗？这是我的代码。

   public class NewClass1 {
        private static int start = 0;
        private static int finish = 0;
        public static int[] time1=new int[20];
        private static  DateFormat sdf = new SimpleDateFormat("yyyy/MM/dd HH:mm:ss");
        private static  DateTimeFormatter dtf = DateTimeFormatter.ofPattern("yyyy/MM/dd HH:mm:ss");
        public static String[] today = new String[4];


        public static NewClass2 myclass(String website)  throws Exception {
            FileWriter fw = new FileWriter("google.txt");
            BufferedWriter bw = new BufferedWriter(fw);
            bw.write("current time                response time");


             NewClass2 getData=new NewClass2();

             DateTimeFormatter dtf = DateTimeFormatter.ofPattern("yyyy/MM/dd HH:mm:ss");

            for(int i=0;i<20;i++)
            {
            String stack_url = website;

            try {
                    URL url = new URL(stack_url);
                     HttpURLConnection httpUrlConnect = (HttpURLConnection) url.openConnection();
                    httpUrlConnect.setConnectTimeout(5000);
                    start = (int) System.currentTimeMillis();
                    httpUrlConnect.connect();

                    if (httpUrlConnect.getResponseCode() == 200) 
                    {
                        finish = (int) System.currentTimeMillis();
                        System.out.println(stack_url + " - " + httpUrlConnect.getResponseMessage() + " took " + (finish-start) + " Milli Seconds.");
                    }

                    if (httpUrlConnect.getResponseCode() == httpUrlConnect.HTTP_NOT_FOUND) 
                    {
                    System.out.println(stack_url + " - " + httpUrlConnect.getResponseMessage() + " - " + httpUrlConnect.HTTP_NOT_FOUND);
                    } 
                } 
                catch (Exception e) 
                {

                }


                 LocalDateTime now = LocalDateTime.now();

                 finish = (int) System.currentTimeMillis();
                 System.out.println(stack_url + " took  "+ (finish-start)+" milliseconds.  at  "+dtf.format(now));
                 String day = dtf.format(now);
                 try{
                 bw.write(day+"             ");
                 bw.write(finish-start);
                 }
                 catch (IOException e) {
                    System.err.println("Problem writing to the file statsTest.txt");
                } 

                  today[i]= dtf.format(now);
                  time1[i]=finish-start;

                    Thread.sleep(6000);
             }
            bw.close();
            fw.close();

            getData.settime(time1);
            getData.settoday(today);

            return getData;

   }

 public static void main(String[] args) throws IOException{
     NewClass1 mynewclass =new NewClass1();

     mynewclass.myclass();

 }
   }

Answer 1

在myclass方法中，它需要一个String参数：

方法myclass

public static NewClass2 myclass(String website)  throws Exception {
    //your code
}

因此，当你调用它时，你应该传入String参数，如下所示：

主要方法

public static void main(String[] args) throws IOException{
    NewClass1 mynewclass =new NewClass1();
    String website = "https://www.google.com";

    MyClass2 yourVariableName = mynewclass.myclass(website);

}

因为你也返回一个变量，所以你应该像这样保持它

Answer 2

myclass函数采用String参数。在main中，您可以使用无参数调用它。

如何将变量从一个方法（有参数）传递给main方法？

2 个答案: