给定字符串如何用数组中的多个单词替换某些单词？答案

【问题标题】：Given string how do I replace certain words with multiple words in an array?给定字符串如何用数组中的多个单词替换某些单词？
【发布时间】：2019-02-17 21:23:10
【问题描述】：

假设我有一个句子和一组单词：

let sentence = 'My * is * years old.';
let words = ['dog', 3];

//expected result => My dog is 9 years old.

如何将每个星号 (*) 替换为给定数组中的相应单词？但我们也可以说存在星号比数组中的元素多的情况：

let sentence = 'The soldiers marched *, *, *, *.';
let words = ['left', 'right'];

//expected result => The soldiers marched left, right, left, right.

使用正则表达式是解决这个问题的唯一方法还是有普通的 JS 解决方案？

【问题讨论】：

你有尝试过什么吗？是的，使用 vanilla js 是可能的。
正则表达式是js内置的，所以是vanilla js。您想出了什么使用正则表达式的解决方案？可以重新调整它的用途，使其不使用正则表达式。
以后，请不要在不符合您的问题的问题上添加标签。我删除了函数式编程标签，因为这与函数式编程无关。您可以将鼠标悬停在标签上以获取其内容的简短描述。了解该网站的许多常规用户都遵循某些标签，对于弹出的与我们的兴趣无关的问题，我们可能会有些脾气暴躁。其次，更不重要的是，正则表达式是 vanilla js :)

标签： javascript arrays regex array-splice

【解决方案1】：

您可以为起始索引使用一个附加值的替换函数，该值的默认值为零。

为了使索引保持在有效范围内，您可以将remainder operator % 与数组的长度一起使用。

const
    replace = (string, array, i = 0) =>
        string.replace(/\*/g, _=> array[i++ % array.length]);

console.log(replace('My * is * years old.', ['dog', 3]));
console.log(replace('The soldiers marched *, *, *, *.', ['left', 'right']));

【讨论】：

我正在考虑从索引的偏移量开始。（除了必要变量的声明和初始化。）

【解决方案2】：

let sentence = 'My * is * years old.';
let words = ['dog', 3];
let count =0;
while (sentence.indexOf('*') > -1) {
    sentence = sentence.replace('*', words[count++]);
    if (count >= words.length) count = 0;
}
console.log(sentence);

显示：“我的狗 3 岁。”

【讨论】：

【解决方案3】：

如果你真的不喜欢正则表达式，你可以这样做：

words = ["left", "right"];
"The soldiers marched *, *, *, *.".split("*").map(function (x, i) {
  return i === 0 ? x : words[(i + 1) % words.length] + x;
}).join("");

执行轨迹：

init  | "The soldiers marched *, *, *, *."
split | ["The soldiers marched ", ", ", ", ", ", ", "."]
i = 0 | ["The soldiers marched ", ", ", ", ", ", ", "."]
i = 1 | ["The soldiers marched ", "left, ", ", ", ", ", "."]
i = 2 | ["The soldiers marched ", "left, ", "right, ", ", ", "."]
i = 3 | ["The soldiers marched ", "left, ", "right, ", "left, ", "."]
i = 4 | ["The soldiers marched ", "left, ", "right, ", "left, ", "right."]
join  | "The soldiers marched left, right, left, right."

【讨论】：

【解决方案4】：

另一个免费的正则表达式选项：

s = "The soldiers marched *, *, *, *.";
f = Function("words", "var i = 0; return \"" + s.split("*").join(
  "\" + words[(i++) % words.length] + \""
) + "\";");

> | f(["L", "R"])
< | "The soldiers marched L, R, L, R."
> | f(["L", "R"].reverse())
< | "The soldiers marched R, L, R, L."
> | f(["L", "R", "R"])
< | "The soldiers marched L, R, R, L."

但要小心恶意代码：

s = "The soldiers marched *, *, *, *.";
s = "\", console.log(\"VIRUS ATTACK!!! CALL +XX-XXXXX-XXXXX NOW!!!\"), \"" + s;
f = Function("words", "var i = 0; return \"" + s.split("*").join(
  "\" + words[(i++) % words.length] + \""
) + "\";");

> | f(["L", "R"])
  | VIRUS ATTACK!!! CALL +XX-XXXXX-XXXXX NOW!!!
< | "The soldiers marched L, R, L, R."

您应该始终清理输入：

s = "The soldiers marched *, *, *, *.";
s = "\", alert(\"VIRUS ATTACK!!! CALL +XX-XXXXX-XXXXX NOW!!!\"), \"" + s;
s = s.split("\"").join("\\\""); // potentially not sufficient!
f = Function("words", "var i = 0; return \"" + s.split("*").join(
  "\" + words[(i++) % words.length] + \""
) + "\";");

> | f(["L", "R"])
< | "", alert("VIRUS ATTACK!!! CALL +XX-XXXXX-XXXXX NOW!!!"), "The soldiers marched L, R, L, R."

但我不敢说它是防弹的：-|

【讨论】：

【解决方案5】：

一个接一个的字符：

input = "The soldiers marched *, *, *, *.";
words = ["left", "right"];
output= ""
for (i = 0, j = 0; i < input.length; i++) {
  output += input[i] !== "*" ? input[i] : (
    words[(j++) % words.length]
  );
}
console.log(output)

如您所见，正则表达式有多种替代方案。无论如何，你真正需要了解的是余数运算符（%）：

0 % 2 = 0 | 0 -> 0
1 % 2 = 1 | 1 -> 1
2 % 2 = 0 | 2 -> 0
3 % 2 = 1 | 3 -> 1
4 % 2 = 0 | 4 -> 0

0, 1, 0, 1, 0, 1, ... 明白了吗？结果永远不会到达正确的操作数。当您需要多次迭代同一个数组时，这尤其有用：

abc = ["A", "B", "C"]
for (i = 0; i < 2 * abc.length; i++) {
  // j will never reach abc.length
  j = i % abc.length; // 0, 1, 2, 0, ...
  console.log(i, j, abc[j]);
}

【讨论】：