你也可以这样做:

Const STR = "hi, there" Const res = str.includes("hello") || str.includes("hi") || str.includes('howdy'); console.log (res);

只要其中一个include返回真值，value就为真，否则，它就为假。这在ES6中工作得非常好。

(错误答案，不要抄)

1线路方案:

字符串/ Array.prototype。包括('hello' || 'hi' || 'howdy');

let words = 'cucumber, mercy, introduction, shot, howdy'
words.includes('hi' || 'howdy' || 'hello') // true
words.includes('hi' || 'hello') // false

这取决于你在什么上下文中使用它。 我在一个对象上使用它来检查是否有任何键有一个空字符串或null作为它的值，它工作

Object.values(object).includes('' || null)

这可以通过使用Array和RegEx的一些/每个方法来完成。

检查list(array)中的所有单词是否存在于字符串中:

const multiSearchAnd = (text, searchWords) => (
  searchWords.every((el) => {
    return text.match(new RegExp(el,"i"))
  })
)

multiSearchAnd("Chelsey Dietrich Engineer 2018-12-11 Hire", ["cle", "hire"]) //returns false
multiSearchAnd("Chelsey Dietrich Engineer 2018-12-11 Hire", ["che", "hire"]) //returns true

检查list(array)中的任何单词是否存在于字符串中:

const multiSearchOr = (text, searchWords) => (
  searchWords.some((el) => {
    return text.match(new RegExp(el,"i"))
  })
)

multiSearchOr("Chelsey Dietrich Engineer 2018-12-11 Hire", ["che", "hire"]) //returns true
multiSearchOr("Chelsey Dietrich Engineer 2018-12-11 Hire", ["aaa", "hire"]) //returns true
multiSearchOr("Chelsey Dietrich Engineer 2018-12-11 Hire", ["che", "zzzz"]) //returns true
multiSearchOr("Chelsey Dietrich Engineer 2018-12-11 Hire", ["aaa", "1111"]) //returns false

那么['hello'， 'hi'， 'howdy'].includes(str)呢?