扫描仪在使用 next() 或 nextFoo() 后跳过 nextLine()?-Java问题

Scanner is skipping nextLine() after using next() or nextFoo()?(扫描仪在使用 next() 或 nextFoo() 后跳过 nextLine()?)

本文介绍了扫描仪在使用 next() 或 nextFoo() 后跳过 nextLine()?的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

我正在使用 Scanner 方法 nextInt() 和 nextLine() 来读取输入.


I am using the Scanner methods nextInt() and nextLine() for reading input. 
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)

问题是输入数值后，第一个input.nextLine()被跳过，第二个input.nextLine()被执行，这样我的输出如下所示:
The problem is that after entering the numerical value, the first input.nextLine() is skipped and the second input.nextLine() is executed, so that my output looks like this:
Enter numerical value
3   // This is my input
Enter 1st string    // The program is supposed to stop here and wait for my input, but is skipped
Enter 2nd string    // ...and this line is executed and waits for my input

我测试了我的应用程序，看起来问题在于使用 input.nextInt().如果我删除它，那么 string1 = input.nextLine() 和 string2 = input.nextLine() 都会按照我的意愿执行.
I tested my application and it looks like the problem lies in using input.nextInt(). If I delete it, then both string1 = input.nextLine() and string2 = input.nextLine() are executed as I want them to be.
推荐答案
那是因为 Scanner.nextInt 方法不会读取通过点击Enter," 所以调用 Scanner.nextLine 在读取该 newline 后返回.

在 Scanner.nextLine 时会遇到类似的行为Scanner.html#next%28%29" rel="noreferrer">Scanner.next() 或任何 Scanner.nextFoo 方法(除了 nextLine 本身).

You will encounter the similar behaviour when you use Scanner.nextLine after Scanner.next() or any Scanner.nextFoo method (except nextLine itself).

解决方法:

在每个 Scanner.nextInt 或 Scanner.nextFoo 之后调用 Scanner.nextLine 以使用该行的其余部分，包括换行

Either put a Scanner.nextLine call after each Scanner.nextInt or Scanner.nextFoo to consume rest of that line including newline

int option = input.nextInt();
input.nextLine();  // Consume newline left-over
String str1 = input.nextLine();

或者，更好的是，通过 Scanner.nextLine 读取输入并将您的输入转换为您需要的正确格式.例如，您可以使用 Integer.parseInt(String) 方法.

Or, even better, read the input through Scanner.nextLine and convert your input to the proper format you need. For example, you may convert to an integer using Integer.parseInt(String) method.

int option = 0;
try {
    option = Integer.parseInt(input.nextLine());
} catch (NumberFormatException e) {
    e.printStackTrace();
}
String str1 = input.nextLine();

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