java - 仅当 ArrayList 等于 bookID 时才更新它吗？

Question

我已经为我的图书馆应用程序设置了一个 Book 类型的 ArrayList。我当前尝试实现的功能是编辑一本书的详细信息。我有一个名为 bookID 的变量，因此当我通过 ArrayList 调用此方法时，它将是 newBook.get(index).getBookID();。根据这些信息，我想检查并执行以下操作:

• 数组中存在具有此 ID 的元素
• 使用新标题更新现有标题

我面临的问题:

• 循环获取ID所在的索引
• 更新现有标题，并将其替换为新标题

到目前为止我想到了什么:

    // Editing book in ArrayList
public void editBook(){
    System.out.println("==============================");
    // Ensuring array has at least one book
    if(newBook.size() > 0){
        // Get which part, and book the user would like to edits
        System.out.print("Enter Book ID to begin editing: ");
        int bookID = sc.nextInt();
        System.out.println("Press 1 or 2 to edit.");
        System.out.println("1: Edit title");
        System.out.println("2: Edit author");
        System.out.print("Enter option: ");
        int editOption = sc.nextInt();

        // Switch statement which will handle editing book
        switch(editOption){
        case 1: 
            // New title
            System.out.print("Enter new title: ");
            String newTitle = sc.nextLine();
            sc.next();

            // Updates title
            newBook.get(position).setTitle(newTitle);

            // Prints out title
            System.out.println(newBook.get(position).getTitle());

            break; // Edit title

上面的代码只是部分代码，下面的任何内容都与问题无关。

Answer 1

直接方法...

所有这些都是循环遍历 Book 的 List 并将用户输入的 bookID 与 bookID<本书的。如果找到匹配项，则 Book 引用将保留在变量 bookToEdit 中。

如果循环完成后，bookToEdit 为 null，则 List 中没有匹配的 Book > 使用指定的 ID，否则，您现在拥有对需要编辑的书籍的引用

// Get which part, and book the user would like to edits
System.out.print("Enter Book ID to begin editing: ");
int bookID = sc.nextInt();
sc.nextLine();
Book bookToEdit = null;
for (Book book : newBook) {
    if (book.getId() == bookID) {
        bookToEdit = book;
        break;
    }
}
if (bookToEdit != null) {

    System.out.println("Press 1 or 2 to edit.");
    System.out.println("1: Edit title");
    System.out.println("2: Edit author");
    System.out.print("Enter option: ");
    int editOption = sc.nextInt();
    sc.nextLine();

    if (editOption == 1 || editOption == 2) {
        System.out.print("New " + (editOption == 1 ? "title: " : "author: "));
        String value = sc.nextLine();
        switch (editOption) {
            case 1:
                // Update title
                break;
            case 2:
                // Update author
                break;
        }
    } else {
        System.out.println("Invalid edit option");
    }

} else {
    System.out.println("Book with the id of " + bookID + " does not exist");
}

流s

如果你想做一些更奇特的事情，你可以使用...

// Get which part, and book the user would like to edits
System.out.print("Enter Book ID to begin editing: ");
int bookID = sc.nextInt();
sc.nextLine();

List<Book> filtered = newBook.stream().filter((Book t) -> t.getId() == bookID).collect(Collectors.toList());
if (!filtered.isEmpty()) {
    Book bookToEdit = filtered.get(0);

现在，您需要知道，这比之前的循环效率低，事实上它将循环遍历 newBook 的整个 List 并返回 ALL与 bookId 匹配的书籍(实际上应该只有一本)

map

最有效的方法是在 Map 中维护 Book，而不是在 List 中，以 为键书号。这样，当需要时，您可以简单地通过 id 查找该书

Map<Integer, Book> newBook = new HashMap<>();
//...
newBook.put(1, new Book(1, "Star Wars", "Somebody"));
newBook.put(2, new Book(2, "Harry Potter", "Somebody else"));
//...

// Get which part, and book the user would like to edits
System.out.print("Enter Book ID to begin editing: ");
int bookID = sc.nextInt();
sc.nextLine();

if (newBook.containsKey(bookID)) {
    Book bookToEdit = newBook.get(bookID);

从概念上讲，这一切都是在生成键和对象之间的映射或关系，这使得根据提供的键更快、更简单地查找对象

看看Collections Trail了解更多详情