Question

我正在尝试设置一个基本示例，我可以扩展我的主程序。我的问题是为什么标题不相等＆＃34; HeyBARK＆＃34;？我把书作为bookTitles课程的一部分，所以不应该自动设置标题添加树皮吗？

完成此操作后，我还需要引用我在原始作业中创建的大小写方法。这个想法将是一旦书名标题设置如下面的代码片段，我将该值传递给方法并将其设置为等于返回结果。如何只访问要修改的标题，因为setter方法将接受整个对象？

class bookTitle {
  constructor(title){
  this.title = title + "BARK";
}

// Everything in comments is part of the second paragraph in the 
// question, I will psuedo code it

// getter the object here to access its title

// use a setter to call my capitalization method to capitalize certain 
// words in the title

// title creator method that returns the modified value, will be called in setter

// cap method that purely caps the correct words, will be called within 
// title creator
}

var book = new bookTitle();
console.log(book); //title is undefinedBARK as expected
book.title = "Hey";
console.log(book); // currently returns Hey, not HeyBARK

如果它有帮助我这里是我当前工作解决方案的实际代码，一个好的开发者能够使它在前一天在一个孤立的环境中工作，但现在我试图修改它以一种setter方式工作。

titleCreator(string) {
        // Note that this isn't meant to be a fully fledged title creator, just designed to pass these specific tests
        var littleWords = ["and", "over", "the"]; // These are the words that we don't want to capitalize

        var self = this; // doesn't need to be here, just for syntax sugar, using this searches for things inside this class

        var modifiedString = string
        .split(' ') // Splits string into array of words, basically breaks up the sentence
        .map(function(word,index) {
            if (index == 0) {
                return self.capitalize(word); // capitalize the first word of the string
            } else if (littleWords.indexOf(word) == -1) {
                return self.capitalize(word); // capitalize any words that are not little, the -1 is returned by indexOf if it can't find the word in the array
            } else if (littleWords.indexOf(word) >= 0) {
                return word; // do not capitalize as this word is in the list of littleWords
            }
        })
        .join(' '); // Joins every element of an array into a string with a space inbetween each value. Basically you created a sentence from an array of words

        return modifiedString;

    }

    capitalize(word) {
        return word.charAt(0).toUpperCase() + word.slice(1);
        // This function just capitalizes the word given to it
    }

Answer 1

book.title = "Hey";

这是在标题中写出值树皮。构造函数在创建时被调用。因此，您执行上述行会替换该值，并且不会重新调用构造函数。

Answer 2

我认为，你在这里误解了构造函数的概念。构造函数构建给定类的新实例，在这种情况下，您将创建BookTitle类的实例。创建新实例后，构造函数不再具有任何权限。使用大写字母和小写实例命名您的类也是一种很好的做法。

要实现您想要的，您需要为您的属性指定getter和setter，例如：

class BookTitle {
  constructor(title){
    this.title = title;
  }
  
  get title() {
    return this._title ;
  }
  
  set title(title) {
    this._title = title + 'BARK';
  }
}

let bookTitle = new BookTitle();
bookTitle.title = 'test';
console.log(bookTitle.title);

Answer 3

您应该在实例化时指定标题，因此您应该使用：

var book = new bookTitle('Hey');

book === 'Hey Bark'

否则，您在实例化后重置整个title属性，因此它忽略了标题在初始化时的设置方式。

为什么在创建实例变量后没有正确设置标题？

3 个答案: