java - 当字符串变量的输入前面有整数输入时，该输入将被跳过

Question

这个问题已经有答案了:

Scanner is skipping nextLine() after using next() or nextFoo()? (25 个回答)

已关闭 3 年前。

我正在练习扫描仪的使用和语法，下面是我编写的代码，我试图从用户那里获取 4 个输入:

    Scanner s= new Scanner(System.in);
    
    int a = s.nextInt();// taking an integer as input, it works fine.
    String b = s.nextLine();//was supposed to take a string input, but not working
    String c=s.nextLine();//takes a string as input from user
    String d=s.nextLine();//takes another string as input from user

    System.out.println("integer entered: "+a);
    System.out.println("first string entered: "+b);
    System.out.println("second string entered: "+c);
    System.out.println("third string entered: "+d);

但是用户只能提供 3 个输入，第一个整数和其他两个字符串输入，并且通过打印这 4 个变量的值，很明显第一个字符串的输入 String b被跳过。然后我尝试重新排列输入的顺序，即

String b = s.nextLine();//takes a string input
int a = s.nextInt();// taking an integer as input
String c=s.nextLine();//was supposed to take a string as input from user
String d=s.nextLine();//takes another string as input from user

在本例中，第二个字符串的输入 String c被跳过。那么，是不是因为整数输入而跳过了下一个字符串输入呢？

Answer 1

这是一个常见问题。

<小时/>

固定代码

Scanner s= new Scanner(System.in);
int a = s.nextInt();
s.nextLine();
System.out.print("Enter a string: ");
String b = s.nextLine();
System.out.print("Enter another string: ");
String c=s.nextLine();
System.out.print("Enter one more string: ");
String d=s.nextLine();

System.out.println(a+"\n"+b+d+c);

引用: Scanner is skipping nextLine() after using next() or nextFoo()?