我为我的库应用程序设置了一个类型为book的ArrayList。我试图实现的当前功能是编辑书的细节。我有一个名为bookID的变量，所以当我通过ArrayList调用此方法时，它将是newBook.get(index).getBookID();。有了这些信息，我想查看，并做以下事情：
我面临的问题：
到目前为止我想出了什么：
// 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
以上代码只是部分代码，低于此代码的任何内容都与问题无关。
答案 0 :(得分：1)
所有这一切都会循环遍历List Book，并将用户输入的bookID与图书的bookID进行比较。如果找到匹配项，则Book引用将保留在变量bookToEdit中。
如果在循环完成后，bookToEdit为null，那么Book中的List与指定的ID不匹配，否则，您现在拥有参考需要编辑的书
// 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");
}
StreamŠ如果你想做一些更有趣的事情，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，这比前一个循环效率低，因为它将遍历整个List newBook，并将返回所有匹配bookId的书籍{1}}（实际上应该只有一个）
Map 最有效的方法是将Book保存在Map中，而不是List，并键入bookId。这样，在需要时，您可以通过它的id
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，所有这一切都是在密钥和对象之间生成映射或关系，这使得根据提供的密钥更快更简单地找到对象
有关详细信息，请查看Collections Trail
答案 1 :(得分：0)