JavaScript 学校练习 - 提示和数组

Question

我需要解决这个学校练习，但我不知道如何解决。正文如下:

我需要制作一个程序，通过 prompt() 对话框从用户那里获取信息。用户需要输入的信息是他的姓名和性别(例如 Marc，m)。基于这些信息，程序需要在 alert() 对话框中写入(已经给出的)列表中的形容词，这些形容词的开头字母与姓名中的字母相同。

例如:用户姓名为Marc，性别为男。程序需要使用男性列表中的形容词(有两个形容词列表，男性和女性)，并在警告对话框中这样写:

疯狂

准确

推理

计算

如果您垂直阅读他们的第一个字母，他们会说 MARC。

我对字母表中的所有字母都有形容词，包括男性和女性。请记住，变量名和形容词是我的母语(塞尔维亚语)，但这应该不是问题，您会明白我的意思，我会在评论中解释代码。

var pridevi = {
  m: ["atraktivan", "blesav", "ciničan", "čudan", "ćopav", "duhovit", "džangrizav", "đavolast", "elokventan", "fantastičan", "grozan", "halapljiv", "imućan", "jak", "katastrofalan", "lep", "ljubazan", "mudar", "naivan", "njanjav", "otporan", "posesivan", "razigran", "smešan", "šaljiv", "tolerantan", "uobražen", "veseo", "zabrinut", "žut"],
  z: ["atraktivna", "blesava", "cinična", "čudna", "ćopava", "duhovita", "džangrizava", "đavolasta", "elokventna", "fantastična", "grozna", "halapljiva", "imućna", "jaka", "katastrofalna", "lepa", "ljubazna", "mudra",  "naivna", "njanjava", "otporna", "posesivna", "razigrana", "smešna", "šaljiva", "tolerantna", "uobražena", "vesela", "zabrinuta", "žuta"],
} // I stored adjectives in object where property m: stands for male and property f: stands for female adjectives

var unos = prompt("Upišite ime i pol. Npr. Mirko, m"); // prompt format

var ime = unos.toLowerCase().split(", ").shift(); // in this variable I stored name
var pol = unos.toLowerCase().split(", ").pop(); // in this variable I stored sex

// console.log(ime + " " + pol) > mirko m

if (unos === null) {
  alert("Korisnik je odustao."); // if user clicks cancel, this message shows in alert dialog
}
else if (unos === undefined && ime < 0 && pol < 0) {
  alert("Nisu uneseni ispravni podaci."); // if user doesn't write the data in correct form, this message shows in alert dialog
}
else {
  var odgovor = pridevi[pol].find(opis => ime[0] === opis[0]); // here's the main thing that doesn't work as it should. it only shows the adjective of the first letter of the name, but not all of them
  alert(odgovor);
}

Answer 1

这个答案依赖于模式匹配，它相当脆弱，因为如果有人不遵循定义的格式，它将无法正常工作。一些注意事项:

• 将形容词组移动到持有对象中，因为访问动态属性相当简单，并且需要更少的 if/else 链接才能到达正确的数组组。
• 定义了一些实用的箭头函数。如果这些不熟悉，请随意使用标准函数尝试它们，如下所示:
  • 函数 notEmpty(val) {
return (val !== undefined && val !== null && val.length > 0);
}
• 用过Array.prototype.map使用回调重新格式化结果数组
• 使用数组解构语法，您可以通过在获取映射结果后定义这些变量来替换它 (const name = mappedArray[0])
• 在 JS 中，您不需要在访问其索引之前拆分字符串，因此 'string'[0] 会为您提供 's'，这在 Array.prototype.find 中很有用回调函数

// adjective group
const adjectives = {
  f: ["accurate", "bodybuilder", "calculative", "decisive" ],
  m: ["abstract", "beautiful", "capable", "delightful" ],
}

// utility function to check that a value exists and isn't empty
const notEmpty = val => (val !== undefined && val !== null && val.length > 0);

// prompt response holder (define a format)
const question = prompt("Write your given name and sex (format: Marc, m.)");

// function to format each entry in the response 
// /\W/g is a regex for stripping non alphanumeric 
const formatEntries = val => val.replace(/\W/g, '').toLowerCase();

// use format to split and handle your response
const [name, adjectiveGroup] = question.split(", ", 2)
  .map(formatEntries)

// if both variables are set then try and find matching adjective
if (notEmpty(name) && notEmpty(adjectiveGroup)) {
    const answer = adjectives[adjectiveGroup].find(adj => name[0] === adj[0])  
  
    console.log(answer);
}