Question

我有一系列名字：

names = [name1, name2, name3...]

对应于包含与名称数组对应的引号的数组数组。

quotes = [[q#1a, q#2b..], [q#1c, q#2d..], [q#1e, q#2f]]

我尝试创建一个可以输入特定引号的函数（例如q#1c），console.log将返回说出来的人的姓名。

我被困住并且一直未定义。有什么建议？以下是我到目前为止的情况。

function findQuote(quote) {
  return quote == `"q#1c"`;
}

let whoSaidIt = (c) => {
  let quote = quotes.index;
  return `${name[c]} said ${quotes[c]}.`;
};

console.log(whoSaidIt(quotes.findIndex(findQuote)));

Answer 1

如果引号是您想要查找的，那么将它们存储在嵌套数组中会使这不必要地变得复杂。我会将引号存储在一个平面数组中并将它们映射到名称。也许是这样的：

＆＃13;

var names = ["Einstein","Trump","Pikachu"];
var quotes = ["E=mcc","Sad.","Pika!","Two things are stupid."];
var quotees = [0,1,2,0];

function whoSaidIt(q) {
  return names[quotees[quotes.indexOf(q)]];
}

document.getElementById("out").innerHTML=whoSaidIt("E=mcc");

＆＃13;

<div id="out"></div>

＆＃13;

Answer 2

let names = ['name1', 'name2', 'name3'];
let quotes = [['q#1a', 'q#2b'], ['q#1c', 'q#2d'], ['q#1e', 'q#2f']];
let whoSaidIt = (c) => quotes.reduce((a, v, i) => quotes[i].includes(c) ? `${names[i]} said ${c}.` : a, '') || 'No-one said that!';

whoSaidIt('q#1c'); //Returns 'name2'

假设names.length === quotes.length和names [I]表示引号[I]，您可以使用reduce来获得答案。一个是＆＃39;＆＃39;默认情况下，由第二个属性决定。

我们只需在找到引号时设置返回消息。我可以用名字来得到说出这句话的人。如果未找到匹配项，我们可以使用||。

保留默认消息

您可能需要考虑使用对象来存储名称和引号，因为它可能更容易引用。 {＆＃39; q＃1a＆＃39;：＆＃39; name1＆＃39;，...}

Answer 3

您的findQuote函数仍然需要查看数组（因为quotes是数组数组），您可以使用includes。此外，如果您将引号传递给搜索而不是在函数中对其进行硬编码，则会更实用。然后，您可以在findQuote：

中使用该参数时将该参数绑定到findIndex

＆＃13;

var names = ['John', 'Helen', 'Anna'];
var quotes = [['q#1a', 'q#2b'], ['q#1c', 'q#2d'], ['q#1e', 'q#2f']];

function findQuote(quote, quotes) {
  return quotes.includes(quote);
}

let whoSaidIt = (c) => {
  return names[c];
};

let quote = "q#1c";
console.log(`${whoSaidIt(quotes.findIndex(findQuote.bind(null,quote)))} said ${quote}`);

＆＃13;

以较短的方式：

＆＃13;

const names = ['John', 'Helen', 'Anna'],
      quotes = [['q#1a', 'q#2b'], ['q#1c', 'q#2d'], ['q#1e', 'q#2f']],
      findQuote = (quote, quotes) => quotes.includes(quote),
      whoSaidIt = c => names[c],
      quote = "q#1c";

console.log(`${whoSaidIt(quotes.findIndex(findQuote.bind(null,quote)))} said ${quote}`);

＆＃13;

如果您可以更改数据结构，那么最好将所有内容放在一起：

＆＃13;

const quotes = [
         { name: 'John', quotes: ['q#1a', 'q#2b'] },
         { name: 'Helen', quotes: ['q#1c', 'q#2d'] },
         { name: 'Anna', quotes: ['q#1e', 'q#2f'] }
      ],
      findQuote = (quote, quotes) => quotes.quotes.includes(quote),
      quote = "q#1c";

console.log(`${quotes.find(findQuote.bind(null,quote)).name} said ${quote}`);

＆＃13;

Answer 4

理想情况下，您的findQuote函数应返回另一个函数。您将引用传递给findQuote，返回findIndex调用数组中每个元素的函数：

function findQuote(quote) {
  return function (el) {
    return el === quote;
  }
}

然后您可以完成代码：

let names = ['Bob', 'Dave', 'Mavis'];
let quotes = ['q#1c','q#2c','q#3c'];

let whoSaidIt = (names, quotes, quote) => {
  const index = quotes.findIndex(findQuote(quote));
  return `${names[index]} said ${quote}`;
};

const quote = 'q#1c';
const result = whoSaidIt(names, quotes, quote);

<强> DEMO

试图在相应数组索引的基础上查找数组中的字符串

4 个答案: