只是想知道，是否有一种方法可以向.includes方法添加多个条件，例如:
var value = str.includes("hello", "hi", "howdy");
想象一下逗号表示“或”。
它现在询问字符串是否包含hello, hi或howdy。所以只有当其中一个条件为真。
有什么方法可以做到吗?
当前回答
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
其他回答
使用includes()，没有，但你可以通过test()实现REGEX:
var value = /hello|hi|howdy/.test(str);
或者，如果词语来自动态来源:
var words = ['hello', 'hi', 'howdy'];
var value = new RegExp(words.join('|')).test(str);
REGEX方法是一个更好的主意，因为它允许您将单词匹配为实际单词，而不是其他单词的子字符串。你只需要边界标记\b这个词，那么:
var str = 'hilly';
var value = str.includes('hi'); //true, even though the word 'hi' isn't found
var value = /\bhi\b/.test(str); //false - 'hi' appears but not as its own word
扩展字符串本机原型:
if (!String.prototype.contains) {
Object.defineProperty(String.prototype, 'contains', {
value(patterns) {
if (!Array.isArray(patterns)) {
return false;
}
let value = 0;
for (let i = 0; i < patterns.length; i++) {
const pattern = patterns[i];
value = value + this.includes(pattern);
}
return (value === 1);
}
});
}
允许你做以下事情:
console.log('Hi, hope you like this option'.toLowerCase().contains(["hello", "hi", "howdy"])); // True
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
那么['hello'， 'hi'， 'howdy'].includes(str)呢?
这是一个有争议的选择:
String.prototype.includesOneOf = function(arrayOfStrings) {
if(!Array.isArray(arrayOfStrings)) {
throw new Error('includesOneOf only accepts an array')
}
return arrayOfStrings.some(str => this.includes(str))
}
允许你做以下事情:
'Hi, hope you like this option'.toLowerCase().includesOneOf(["hello", "hi", "howdy"]) // True
