Question

我有一个外部文本文件：

-To Kill a Mockingbird by Harper Lee.
-The Great Gatsby by Scott Fitzgerald.
-Hamlet by William Shakespeare.
-Then Catch in the Rye by J.D Salinger.
-One Hundred Years of Solitude by Gabriel Garcia Marquez.
-The Hobbit by J.R.R Tolkien.
-Moby Dick by Herman Melville.
-A Tale of two Cities by Charles Dickens.
-Don Quixoteby Miguel de Cervantes.
-Love in the Time of Cholera by Gabriel Garcia Marquez.
-Of Mice and Men by John Steinbeck.
-Fahrenheit 451 by Ray Bradbury.
-Stranger in a Strange Land by Robert Heinlein.
-Siddartha by Herman Heese.
-Atlas Shrugged by Ayn Rand.
-The Count of Monte Cristo by Alexandre Dumas.
-The Iliad by Homer.
-The Odyssey by Homer.
-A Wrinkle in Time by Madeleine L'Engle.
-Inferno by Dante Alighieri.
-Paradise Lost by John Milton.
-Alice's Adventures in Wonderland by Lewis Carroll.
-War and Peace by Leo Tolstoy.
-Frankenstein by Mary Shelley.
-Romeo and Juliet by William Shakespeare.
-Exodus by Leon Uris.
-1984 by George Orwell.

我想要做的是分割每一行的字符串并将它们存储在一个arraylist上我只是不知道为什么它在读取这个文本文件时从第一行跳到第三行：我的代码：

bookSearch = new Scanner(new FileInputStream("src/booksNames.txt")).useDelimiter(" by ");
            books = new ArrayList<Books>();
            String storeName = "";
            String storeAuthor = "";

            while(bookSearch.hasNextLine())
            {

                storeName = bookSearch.next().split("by")[0];
                storeAuthor = bookSearch.next().split("(by)|(\\.)")[0];

                bookSearch.nextLine();              

                info = new Books(storeName, storeAuthor);
                books.add(info);
            }

我得到的是由Harper Lee杀死一只嘲笑鸟，然后由William Shakespeare跳到哈姆雷特！它只是忽略了第二，第四，第六行等等..... 任何帮助将不胜感激!!

每个标题和作者都是一个单独的行！

Answer 1

你在loop.bookSearch.nextLine（）里面两次调用bookSearch.next（），每次都跳到下一个对象。

while(bookSearch.hasNextLine())
            {

                storeName = bookSearch.next().split("by")[0]; 
                storeAuthor = bookSearch.next().split("(by)|(\\.)")[0]; // The error lies here, bookSearch.next() skips to the next object every time

                bookSearch.nextLine();              

                info = new Books(storeName, storeAuthor);
                books.add(info);
            }

正确实施：

while(bookSearch.hasNextLine())
            {
                String bookString = bookSearch.nextLine();
                storeName = bookString.split("by")[0];
                storeAuthor = bookString.split("by")[1];

                info = new Books(storeName, storeAuthor);
                books.add(info);
            }

Answer 2

这是因为您在bookSearch.next()循环

中两次致电while

Answer 3

您已在循环内调用bookSearch.next()两次，并在bookSearch.nextLine()调用一次。

查找并返回此扫描仪的下一个完整令牌。在完成令牌之前和之后是与分隔符模式匹配的输入。即使之前的hasNext（）调用返回true，此方法也可能在等待输入扫描时阻塞。

while(bookSearch.hasNextLine())
{
      String[] book = bookSearch.next();
      storeName = book.split("by")[0];
      storeAuthor = book.split("(by)|(\\.)")[1];
      info = new Books(storeName, storeAuthor);
      books.add(info);
}

Answer 4

使用类似的东西

try{
        FileReader fr=new FileReader("Test.txt");
        BufferedReader br=new BufferedReader(fr);

        while((str=br.readLine()) != null){
        strBuf.append(str);
                //here you can add the str to your arrayList

        }
        }catch(Exception e){

        }

此处Test.txt是您的输入文件。此代码段将逐行获取

来自文本文件。

希望这就是你要找的东西

Answer 5

您可以使用以下代码来读取文件。您还应该考虑以下事项。

您应首先检查它是否包含“by”关键字，因为有一行不包含“by”，例如“Don Quixoteby Miguel de Cervantes。”。

你应该使用“by”关键字之间的空格进行拆分，因为“Gatsby”也包含“by”，它也会拆分它并给你错误的结果。

public void readFile(String fileName) {
DataInputStream in = null;
try {
    in = new DataInputStream(new FileInputStream(new File(fileName)));
    BufferedReader br = new BufferedReader(new InputStreamReader(in));

    String line;
    while ((line = br.readLine()) != null) {
        if (line.contains(" by ")) {
            String[] arr = line.split(" by ");
            String book = arr[0];
            String author = arr[1];

            System.out.println("Book : " + book + " Author : " + author + "\n");
            System.out.println("\n");
        } else {
            System.out.println(line + "\n");
        }
    }
} catch (UnsupportedEncodingException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
} finally {
    if (in != null) {
        try {
            in.close();
        } catch (IOException e) {
            e.printStackTrace();
        }
    }
}
}

Answer 6

那里你大概有90％。需要注意的一点是String.split()返回的内容 - 它会返回一个String数组，您可以正确记录它。但是，如果您为正在阅读的每一行创建数组并以此方式执行处理，那么您将获得更好的服务。

此外，您正在拨打next()两次。回想一下Scanner.next()实际上做了什么 - 最好是创建局部变量来保存String或分割String[]。

拆掉这个字符串？

6 个答案: