Question

例如，用户可以在我的程序中输入这样的输入：
function onNewBookMarkCallBack(){ $('#idOfTheItemThatYouWantToChangeTheClass').addClass("newBookMark"); } function onBookMarkDeletedCallBack(){ $('#idOfTheItemThatYouWantToChangeTheClass').removeClass("newBookMark"); }

或者像这样：

71 117 48 115 127 125 117 48 121 126  48  96 117 113 115 117

用户只能通过连续按两次“Enter”来终止输入流。

我该怎么做？

Answer 1

您可能希望更改从hasNextInt()到hasNextLine()的输入方式以及nextInt()到nextLine()

的输入方式

boolean enterOnce = false;

while(scanner.hasNextLine()) {
    String line = scanner.nextLine();

    if(line.isEmpty())
        if(enterOnce)
            break;
        else
            enterOnce = true;
}

Answer 2

hasNextInt和nextInt忽略空格;所以你不能使用它们。您可以使用hasNextLine和nextLine（前一行存储）;然后解析每个输入行的值（停在两个空行）。此外，您可以使用菱形运算符<>（我建议编程到List接口（而不是具体的ArrayList实现）。像

这样的东西

public static void main(String[] args) {
    List<Integer> integers = new ArrayList<>();
    Scanner scanner = new Scanner(System.in);
    String prevLine = "";
    String line = null;

    while (scanner.hasNextLine()) {
        if (line != null) {
            prevLine = line;
        }
        line = scanner.nextLine().trim();
        if (line.isEmpty() && prevLine.isEmpty()) {
            break;
        }
        String[] parts = line.split("\\s+");
        for (String p : parts) {
            integers.add(Integer.parseInt(p));
        }
    }
}

Answer 3

我发现以前的答案没有按要求进行。一种解决方案是在总共两个空行而不是两个连续空行之后退出循环。

2 4
<<ENTER>>
543
<<ENTER>>
---loop breaks---

如果该空行是第一行，则另一种解决方案在单个空行之后退出：

<<ENTER>>
---loop breaks---

下面，我以不同的方式实现对连续空行的跟踪，以便正确处理这两种情况。此外，为了防止 InputMismatchExceptions，我还在将每个标记添加到整数列表之前验证它是一个整数。

public static void main(String[] args) {
     
    // Initialise list to store the integers from input.
    List<Integer> integers = new ArrayList<>();

    // Initialise keyboard stream to get integers from keyboard.
    Scanner inputStream = new Scanner(System.in);

    // Declare a scanner to parse the lines read in from the inputStream.
    Scanner lineReader;

    // Initialise boolean to track whether two consecutive ENTER keys
    // have been pressed.
    boolean previousLineEmpty = false;
    
    // Continue taking input from the user and adding the integer input to
    // a list until ENTER is pressed twice.
    while (inputStream.hasNextLine()) {
        
        String line = inputStream.nextLine();
        
        // Determine whether the loop should break or add the integers in 
        // the line to the list.
        if (line.isEmpty()) {
          
            // If the current line and previous line are empty,
            // ENTER was pressed twice. So break.
            if (previousLineEmpty) {
                break;
            }
            
            // Otherwise, this line is empty and is an empty previous line for 
            // the next iteration.
            else {  
                previousLineEmpty = true;
            }
            
        } else {
            
            // Initialise scanner to process tokens in line.
            lineReader = new Scanner(line);
            
            // Process the tokens in the non-empty line, adding integers to the
            // list and ignoring non-integers.
            while (lineReader.hasNext()) {

                // Add token to list if it is an integer.
                if (lineReader.hasNextInt()) {
                    integers.add(lineReader.nextInt());
                } 
                // If token is not an integer, instead advance to next token.
                else {
                    lineReader.next();
                }
            }   
            
            // In the next iteration, this non-empty line is the previous 
            // line. Set boolean to false.
            previousLineEmpty = false;
        }  
    }

如何在连续按两次“Enter”键时终止`System.in`键盘流？

3 个答案: