相关疑难解决方法(0)

在使用next()或nextFoo()之后,Scanner正在跳过nextLine()?

我正在使用这些Scanner方法nextInt()nextLine()阅读输入.

它看起来像这样:

System.out.println("Enter numerical value");    
int option;
option = input.nextInt(); // Read numerical value from input
System.out.println("Enter 1st string"); 
String string1 = input.nextLine(); // Read 1st string (this is skipped)
System.out.println("Enter 2nd string");
String string2 = input.nextLine(); // Read 2nd string (this appears right after reading numerical value)
Run Code Online (Sandbox Code Playgroud)

问题是输入数值后,第一个input.nextLine()被跳过而第二个input.nextLine()被执行,所以我的输出如下所示:

Enter numerical value
3   // This is my input
Enter 1st string    // The program is supposed to stop here and …
Run Code Online (Sandbox Code Playgroud)

java io java.util.scanner

627
推荐指数
17
解决办法
45万
查看次数

标签 统计

io ×1

java ×1

java.util.scanner ×1