它到达if语句并停止工作，我在做什么（java新）[duplicate]

Question

import java.util.Scanner;

public class JavaMovies {
      Scanner det = new Scanner(System.in);
      int age = det.nextInt();

public static void main(String[] args){
    calculator();


}


public static void calculator(){

Scanner det = new Scanner(System.in);

    String rMovie = "Attack of the nerds";
    String mMovie = "Nerd lyfe";
    String pgMovie = "Nerd adventures";
    String gMovie = "Nerd playtime";
    System.out.println("These are the movies that are playing this weekend"
            + rMovie
            + mMovie
            + pgMovie
            + gMovie);
    System.out.println("Please input your age");
   int age = det.nextInt();


System.out.print("Input the movie you wish to see:");
    String x = det.nextLine();                             //Program ends here...

    if(x.equals("Attack of the nerds") & age > 17){
        System.out.println("You can see Attack of the nerds");

    }else if(x.equals("Attack of the nerds") & age < 18){
        System.out.println("You can see attack of the nerds!");

    }if(x.equals("Nerd lyfe")& age > 15){
        System.out.println("You can see Nerd lyfe");

    }else if(x.equals("Nerd lyfe")& age < 16){
        System.out.println("You can not see Nerd lyfe");

    }if (x.equals("Nerd adventures")& age > 10){
        System.out.println("You can see Nerd adventures");

    }else if(x.equals("Nerd adventures")& age < 11){
        System.out.println("You can not see Nerd adventures");

    }if(x.equals("Nerd playtime")& age > 5){
        System.out.println("You can see Nerd playtime");

    }else if(x.equals("Nerd playtime")& age < 6){
        System.out.println("You can not see Nerd playtime");


    }
}
}

在if语句开始之前一切正常，我可以做什么来修复这个问题呢？

Answer 1

我认为问题是nextint没有用完整行。如果在提示输入年龄时键入32，nextint将使用3和2字符，但输入字符串中仍保留一个换行符。然后，当您调用nextline时，效果是返回当前行中剩余的所有内容，直到下一个换行。由于当前行仍有换行符，这将导致字符串x=det.nextLine()将x设置为空字符串。

在提示播放电影之前，请尝试det.nextline()；。（您不需要将其赋值给变量；只需将结果丢弃。）这将导致扫描程序消耗换行符，以便下一个nextline()将获得另一行输入。