Question

嗨，我想用JavaScript创建一个非常基本的亵渎过滤器。

我有一个名为badWords的数组，也有一个常量叫description。我不会检查该description中是否包含任何不良词。

这是我到目前为止所做的。

const badWords = ["Donald Trump","Mr.Burns","Sathan"];

const description = "Mr.Burns entered to the hall."
let isInclude = false;
badWords.forEach(word=>{
  if(description.includes(word)){
  isInclude = true
  }
})

console.log(`Is include`,isInclude)

唯一的问题是我必须遍历badWords数组。有没有一种方法可以在不循环遍历数组的情况下完成此任务？

Answer 1

使用some()-一旦找到条件匹配项，它就会从循环中退出，因此它比循环更有效。

let isInclude = badWords.some(word => description.includes(word));

Answer 2

这是正则表达式解决方案的样子：

// https://stackoverflow.com/a/3561711/240443
const reEscape = s => s.replace(/[-\/\\^$*+?.()|[\]{}]/g, '\\$&');

// needs to be done only once
const badWords = ["Donald Trump","Mr.Burns","Sathan"];
const badWordsRE = new RegExp(badWords.map(reEscape).join('|'));

// testing is quick and easy
console.log("Mr.Burns entered to the hall.".match(badWordsRE)); // "Mr.Burns"
console.log("Nothing objectionable".match(badWordsRE));         // null

（如果您的坏话是实际的正则表达式，例如"Mr\.Burns"，则省略.map(reEscape)）

Answer 3

使用try catch

const badWords = ['Donald Trump', 'Mr.Burns', 'Sathan']

const description = 'Mr.Burns entered to the hall.'
let isInclude = false
try {
  badWords.forEach(word => {
    if (description.includes(word)) {
      isInclude = true
      throw new Error(word)
    }
  })
} catch (e) {
  console.log(e)
}

JavaScript简单亵渎过滤器

3 个答案: