Question

所以我有这个代码，我有一个句子数组，通过一个开关提示用户要询问他们想要显示多少句子1 - 5.然后它将循环并随机显示的数量从阵列中请求的句子。

＆＃13;

var sentences = ["sentence one", "sentence two", "sentence three",
  "sentence 4", "sentence 5"];

function shuffle(a) {
  for (i = a.length - 1; i; i--) {
    let j = Math.floor(Math.random() * i);
    [a[i - 1], a[j]] = [a[j], a[i - 1]];
  }
}

var random = prompt("how many sentences do you want?");
switch (random) {
  case "1":
    shuffle(sentences)
    console.log(sentences(0, 1));
    break;

  case "2":
    shuffle(sentences)
    console.log(sentences.slice(0, 2))
    break;

  case "3":
    shuffle(sentences)
    console.log(sentences.slice(0, 3))
    break;

  case "4":
    shuffle(sentences)
    console.log(sentences.slice(0, 4))
    break;

  case "5":
    shuffle(sentences)
    console.log(sentences.slice(0, 5))
    break;

  default:
    console.log("incorrect number");
}

＆＃13;

代码工作正常，但是当它返回时，它会显示当然的数组：如果这个人想要3个句子：

(3) ["sentence two", "sentence four", "sentence five"]

我想返回句子本身而不是数组：

sentence two, sentence four, sentence five

我尝试使用.length转换为字符串，但这会返回字符串中的字符数，例如输入的提示为5，因此返回：

sente

任何人都有任何想法，非常感谢：）

Answer 1

在显示之前使用.slice去除括号。

https://www.w3schools.com/jsref/jsref_slice_array.asp

https://www.w3schools.com/jsref/jsref_slice_string.asp

或者你可以使用split：

https://www.w3schools.com/jsref/jsref_split.asp

或者您可以使用替换：

https://www.w3schools.com/jsref/jsref_replace.asp

Answer 2

你可以通过＆＃34;加入＆＃34;你的数组有join method。例如，给定["foo", "bar"]数组，调用["foo", "bar"].join(", ")将输出包含foo, bar的字符串。

以下是一个例子：

> var foo = ['bar', 'bat'];
[ 'bar', 'bat' ]
> foo.join(', ');
'bar, bat'

Answer 3

要实现您的目标，您可以在ES6中使用传播...运算符。

var sentences = ["sentence one", "sentence two", "sentence three", 
"sentence 4", "sentence 5"];

  function shuffle(a) {
        for (i = a.length - 1; i; i--) {
            let j = Math.floor(Math.random() * i);
            [a[i - 1], a[j]] = [a[j], a[i - 1]];
        }
    }

 var random = prompt("how many sentences do you want?");
  switch (random) {
    case "1":
    shuffle(sentences)
   console.log(...sentences(0, 1));
      break;

      case "2":
       shuffle(sentences)
      console.log(...sentences.slice(0, 2))
      break;

      case "3":
       shuffle(sentences)
      console.log(...sentences.slice(0, 3))
      break;

      case "4":
       shuffle(sentences)
      console.log(...sentences.slice(0, 4))
      break;

      case "5":
       shuffle(sentences)
      console.log(...sentences.slice(0, 5))
      break;

      default:
      console.log("incorrect number");
}

作为旁注，switch语句是重复且不必要的：

 var random = +prompt("how many sentences do you want?");
 if (0 <= random <= 5) {
   shuffle(sentences);
   console.log(...sentences.slice(0, random));
 } else {
   console.log("invalid number");
 }

这里，+将数字字符串转换为数字。您还可以使用Number(prompt(..))显式转换。

编辑：我看到现在你想在元素之间插入逗号。

您可以使用Array.prototype.join。

 var random = +prompt("how many sentences do you want?");
 if (0 <= random <= 5) {
   shuffle(sentences);
   console.log(sentences.slice(0, random).join(', ');
 } else {
   console.log("invalid number");
 }

Answer 4

使用join()将数组中的所有值与字符串连接起来（如', '）

＆＃13;

var sentences = ["sentence one", "sentence two", "sentence three",
  "sentence 4", "sentence 5"];

function shuffle(a) {
  for (i = a.length - 1; i; i--) {
    let j = Math.floor(Math.random() * i);
    [a[i - 1], a[j]] = [a[j], a[i - 1]];
  }
}

var random = prompt("how many sentences do you want?");

shuffle(sentences);

switch (random) {
  case "1":
    sentences = sentences.slice(0, 1);
    break;

  case "2":
    sentences = sentences.slice(0, 2);
    break;

  case "3":
    sentences = sentences.slice(0, 3);
    break;

  case "4":
    sentences = sentences.slice(0, 4)
    break;

  case "5":
    sentences = sentences.slice(0, 5);
    break;

  default:
    console.log("incorrect number");
}

console.log(sentences.join(', '));

＆＃13;

Answer 5

您可以使用join ()。 join（）方法将数组的所有元素（或类数组对象）连接成一个字符串并返回该字符串。

在您的代码中：

 console.log((sentences.slice(0, 3)).join(', '));

如何在没有括号javascript的情况下返回数组

5 个答案: