我的java代码没有错误，但是为什么赢了'；它不跑吗？

出于某种原因，当我在eclipse上运行代码时，屏幕上没有显示出应有的内容。我不知道是什么。我可以在控制台中键入输入，但仅此而已。没有别的事情发生。有人能告诉我可能是什么吗？谢谢

import java.util.Scanner;

public class Test {

public static void main (String [] args) {
    Scanner input = new Scanner (System.in);

    Flight flight = new Flight (input.nextLine(), input.nextLine(), input.nextLine(), input.nextLine(), 0);

    System.out.println("Creating first flight");
    System.out.println("What is the name of the flight?");
        String flightName = input.nextLine();
    System.out.println("What is the destination of the flight?");
        String destination = input.nextLine();
    System.out.println("What is the departure time of the flight?");
        String departureTime = input.nextLine();
    System.out.println("What is the departure gate of the flight?");
        String departureGate = input.nextLine();

    boolean done = false;
    while (!done) {
        System.out.println("Now what would you like to do?");
        System.out.print("1. Print out a flight's info");
        System.out.print("2. Print out the number of flights through the static variable.");
        System.out.print("3. Change the departure time of a flight.");
        System.out.print("4. Change the departure gate of a flight.");
        System.out.print("5. Exit");
        int choice = input.nextInt();

        switch (choice) {

        case 1: 
            System.out.println("Which flight would you like to print the info of (1 or 2)?");
                int selection = 0;
                selection = input.nextInt();
            if (selection == 1 || selection == 2) {
                Flight.printFlight();
            } else{
                System.out.println("Invalid Option");
            } break;

        case 2: 
            System.out.println("This is the number of flights" + Flight.getNumFlights());
            break;

        case 3: 
            System.out.println("Which flight would you like to change the departure time of (1 or 2)?");
                int selection2 = 0;
                selection2 = input.nextInt();
            if (selection2 == 1 || selection2 == 2){
                System.out.println("What is the new departure time for flight " + (Flight.getNumFlights()-1));
                    String newDeptTime = input.nextLine();
                    Flight.changeDeptTime(newDeptTime);
            } else{
                System.out.println("Invalid Option");
            } break;

        case 4: 
            System.out.println("Which flight would you like to change the departure gate of?");
                int selection3 = input.nextInt();
            if (selection3 == 1 || selection3 == 2){
                System.out.println("What is the new departure gate for flight " + Flight.getNumFlights());
                    String newDeptGate = input.nextLine();
                    Flight.changeDeptGate(newDeptGate);
            } else {
                System.out.println("Invalid option");
            } break;

        case 5: 
            done = true;
            break; 
        default:
            System.out.println("Invalid option");
            break;

        }

    }   

}

}

---

它没有显示任何内容，因为它正在等待您在Flight class构造函数中输入所需的字符。
正如你在这里看到的

Flight flight = new Flight (input.nextLine(), input.nextLine(), input.nextLine(), input.nextLine(), 0);

执行在

input.nextLine（）

parts处停止，直到输入内容并按enter键

---

附言：最好放置一些

System.out.println（“输入值”）

在输入部分之前的一种语句，用于向用户显示要执行的操作。有趣的是，这样做可以防止您犯此错误

在调试中插入断点并运行它，然后逐步检查代码并查看其中断位置。即使为

航班

，对吗？除非您输入一些文本，否则不会显示任何内容，因为您在构建航班时要求使用扫描仪在标准输入上输入。哎呀……我的眼睛完全错过了……谢谢everyone@user7087153这就是调试器应该能够在几秒钟内告诉您的。