Question

通常，我在扫描某些元素的数组列表时没有问题。我知道如何构建while循环等。但是，在这种情况下，我需要使用扫描仪，但它给出了我的问题，如下所示：

enter image description here

以下代码旨在使用扫描程序输入作者和标题，以检查该精确书籍（包括作者和标题的精确匹配）是否在数组列表中。

我很可能忽略了一些简单的事情，但无论如何，我不需要任何评论来评论这是一个愚蠢的代码等。

public String checkForBookUsingInfo(){
    int index = 0;
    Book bookObject = null;
    String returnValue = "Book not found";
    String title = "";
    String author = "";
    Boolean isFound = false;
    while (index <bookList.size() && isFound == false ){
        bookObject = bookList.get(index);
        System.out.println("Please enter title of book to search for.");
        String anyTitle = keybd.next();
        System.out.println("Please enter author of book to search for.");
        String anyAuthor = keybd.next();
        if ((title.equals(anyTitle)) && (author.equals(anyAuthor))){
            returnValue = "Book is in library.";
        }
        index++;
    }
    return returnValue;

Answer 1

next()只返回一个令牌（字），因此对于像The Prince这样的数据，首先next()会返回"The"秒next()将返回"Prince" （因此它不会等待来自用户的输入，因为它已经有了它的标记）。

如果您想阅读多个单词，请使用nextLine()读取整行。

如果您想在next()和nextLine()代码中使用，请阅读Scanner is skipping nextLine() after using next(), nextInt() or other nextFoo() methods

还有其他一些问题：

如果找到图书，则不会将isFound设置为true;
每次迭代书时你都要求书名和作者，但你应该在迭代之前知道这些信息，所以也许让用户将这些信息作为方法参数传递。
您正在将用户提供的值与来自""和title字段的空字符串（author）进行比较

您的代码可能看起来更像：

class Book{
    private String author;
    private String title;

    //getters and setters
}


class Library {

    private List<Book> bookList = new ArrayList<Book>();

    public String checkForBookUsingInfo(String author, String title){
        for (Book book : bookList){
            if (book.getAuthor().equals(author) && book.getTitle().equals(title)){
                return "Book is in library.";
            }
        }
        return "Book not found in library";
    }

    public static void main(String[] args) throws Exception {
        Scanner keybd = new Scanner(System.in);
        Library library  = new Library();
        //add some books to library
        //....

        System.out.println("Please enter title of book to search for.");
        String anyTitle = keybd.nextLine();

        System.out.println("Please enter author of book to search for.");
        String anyAuthor = keybd.nextLine();

        String stateOfBook = library.checkForBookUsingInfo(anyAuthor, anyTitle);
        System.out.println(stateOfBook);

    }
}

Answer 2

事实证明，其他答案让我思考（我今天还没有完成）。显然我正在寻找我想要的东西，我只需要重新排列东西。虽然下面的代码有效，但另一种方法更容易阅读。

public String checkForBookUsingInfo(){
    int index = 0;
    Book bookObject = null;
    String returnValue = "Book not found";
    String title = "";
    String author = "";
    Boolean isFound = false;
    System.out.println("Please enter the name of a book to search for.");
    String anyTitle = keybd.nextLine();
    System.out.println("Please enter the name of an author to search for.");
    String anyAuthor = keybd.nextLine();
    while (index <bookList.size() && isFound == false ){
        bookObject = bookList.get(index);
        title = bookObject.getTitle();
        author = bookObject.getAuthor();
        if ((title.equals(anyTitle)) && (author.equals(anyAuthor))){
            returnValue = "Book is in library.";
        }
        index++;
    }
    return returnValue;
}

使用Scanner和While循环来检查arraylist

2 个答案: