java扫描器不会停止获取输入

Question

好的，非常好的问题。我正在制作一个CLI应用程序，允许用户设计调查。首先他们输入问题，然后输入选择的数量和选择。我使用扫描仪获取输入，出于某种原因，它允许用户输入大部分内容，但不允许输入问题的文本。下面是代码片段

String title = "";
Question[] questions;
int noOfQuestions = 0;
int[] noOfChoices;
Scanner entry = new Scanner(System.in);
System.out.println("Please enter the title of the survey: ");
title = entry.nextLine();
System.out.println("Please enter the number of questions: ");
noOfQuestions = entry.nextInt();
noOfChoices = new int[noOfQuestions];
questions = new Question[noOfQuestions];
for (int i = 0; i < noOfQuestions; i++) {
    questions[i] = new Question();
}
for (int i = 0; i < noOfQuestions; i++) {

    System.out.println("Please enter the text of question " + (i + 1) + ": ");
    questions[i].questionContent = entry.nextLine();
    System.out.println("Please enter the number of choices for question " + (i + 1) + ": ");
    questions[i].choices = new String[entry.nextInt()];
    for (int j = 0; j < questions[i].choices.length; j++) {
        System.out.println("Please enter choice " + (j + 1) + " for question " + (i + 1) + ": ");
        questions[i].choices[j] = entry.nextLine(); 

    }
}

谢谢：）

Answer 1

我询问您是否从扫描仪读取noOfQuestions的原因是Scanner.nextInt()不使用分隔符（例如新行）

这意味着下一次调用nextLine()时，您将从上一个readInt()中得到一个空字符串

75\nQuestion 1: What is the square route of pie?
^ position before nextInt()

75\nQuestion 1: What is the square route of pie?
  ^ position after nextInt()

75\nQuestion 1: What is the square route of pie?
    ^ position after nextLine()

我的建议是逐行阅读，总是使用nextLine()，然后使用Integer.parseInt()进行解析

如果这是你的路线，根本不需要扫描仪；您可以只接受一个BufferedReader

Answer 2

{a1}的文档说明将扫描器前进到当前行，并返回跳过的输入可能解释了您看到的行为。只是为了验证您是否可以添加一个sysout并在title = entry.nextLine()之后打印title的值，并查看它所包含的值

如果要从输入中读取完整的行，可能需要使用InputStreamReader combination with BufferedReader