java Piglatin，难以理解如何移动“.”一串

Question

我在为我的计算机科学课写的这个程序上遇到了很大的困难。我浏览了学生们在这里和其他网站上发布的其他课程，但我一直无法理解我做错了什么。我已经看过JavaOracle文档，但它并没有让我满意

我的主要问题是嵌套的部分if语句，它应该移动“.”在单词后面加上“ay”后，放在句末。应用程序不完整。在我学会如何解决这一步后，我将继续讨论句子中第一个单词的大写。我只是不明白为什么我在这方面有困难。子字符串和。查拉特

请解释一下答案

    File input = new File ("elab.txt");
    File output = new File ("elab.out");

    //create Scanner
    Scanner in = new Scanner(input);

    //create writer
    PrintWriter out = new PrintWriter(output);

    //VariParty, Part 1
    int count = 0;//Simple counter
    String word;//Scanned word from text file, will be edited
    String original;//Original word, backed up from String "word" before edit
    String original1;//Used in consonants
    String sentence = "";//Declare full sentence

    /* What I will do is scan each word in the text document. The application will display
     * a number for each word for my visual learning pleasure. The counter will be //commented
     * out as to not cause any visual action for the complete application. 
     */


    while(in.hasNext()){//Star while
        word = in.next();//Scans individual word
        word = word.toLowerCase();
        count++;//adds to counter
        original = word.toLowerCase();
        original1 = word.toLowerCase();

        //VariParty, Part 2
        char fl = word.charAt(0);
        char fL = Character.toLowerCase(word.charAt(0));
        char endChar = word.charAt(word.length()-1);


        /* This vowel portion of the application will check each "word" and deduce if "ay"
         * should be added to the end of the word beginning with a vowel. Finally, the code
         * will build onto the sentence string to produce a sentence.
         */

        if (fl == 'a' || fl == 'e' || fl == 'i'|| fl == 'o' || fl == 'u'){//Start if
        word = word + "ay";
        System.out.println(count + " VOW) " + "\"" + original + "\"" + " to " + "\"" + word + "\"" + " ");
        out.print(word + " ");                      
        sentence = sentence + " " + word;           
        }//End if


        /* We will now work towards building a better sentence. Our sentence will not have the
         * audacity to place a period in the middle of the word! Our sentences shall forever be
         * properly defined with a punctuation mark at the end, unless it's an apostrophe.              
         */

            if(word.contains(".")){//Start nested if
                word = word.substring(0,word.length()-1);
                word = word + ".";
                System.out.println(count + " NES VOW) " + "\"" + original + "\"" + " to " + "\"" + word + "\"" + " ");
            }//End nested if


        /* Now we will move the word around and deal with the consonants
         * The else statement will build the proper word by moving the first
         * letter to the end of the sentence, then adding "ay" to the end.
         * The final line will be building onto the sentence string.
         */


        else{//Start else
            word = word.substring(1,word.length());
            original = word;
            word = word + fl + "ay";
            System.out.println(count + " CON) " + "\"" + original1 + "\"" + " to " + "\"" + word + "\"" + " ");
            out.print(word + " ");

            sentence = sentence + " " + word;

        }//End else 
    }//End While

    System.out.println(sentence);
    in.close();//Close Scanner
    out.close();//Close PrintWriter 

Answer 1

在下面的一行中，您的结束索引是单词的结束。因此，我不确定您在这一步中想做什么

word = word.substring(0,word.length()-1);

根据文件

public String substring(int beginIndex,
                    int endIndex)

因此，你的步骤应该保持单词不变。这是你的问题吗。我不能评论，所以添加这个作为答案

另外，如果要在while循环中修改字符串，请尝试使用Stringbuffer

Answer 2

您应该做的第一件事是使您的程序可测试

主要功能是将单词转换成拉丁语。因此，您应该定义一种方法，该方法可以准确地执行以下操作：

public static String translateWordToPigLatin(String word) {
    ....
    return ...;
}

然后，您应该编写另一个名为PigLatinTest的类，在其中测试此方法：

public class PigLatinTest {

    public static void main(String[] args) {
        testPigLatin("hello", "ellohay");
        testPigLatin("welcome", "ellcomeway");
        // Add more test cases here.
    }

    private static void testPigLatin(String english, String expectedPigLatin) {
        String actual = translateWordToPigLatin(english);
        if (!expectedPigLatin.equals(actual)) {
            throw new AssertionError("Expected \"" + expectedPigLatin + "\" for \"" + english + "\", got \"" + actual "\".");
        }
    }
}

在translateWordToPigLatin方法中，我调用了参数word。这是一个强有力的迹象，表明“.”不属于那里。所以你真正应该做的是从这行中提取所有合适的单词，其中一个单词只由字母组成

下一个更大的方法应该称为translateLineToPigLatin(String line)，在这个方法中，您应该：

1. 从英语行的开头开始
2. 将字符复制到pig拉丁行，直到找到一个字母
3. 进一步扫描英文行，直到你再也找不到字母
4. 步骤2和步骤3之间的文本是一个单词
5. 把那个词转换成拉丁语
6. 将转换后的单词添加到pig拉丁语行
7. 继续扫描你停下来的英语行