multiple conditions for JavaScript .includes() method

Just wondering, is there a way to add multiple conditions to a .includes method, for example:

    var value = str.includes("hello", "hi", "howdy");

Imagine the comma states "or".

It's asking now if the string contains hello, hi or howdy. So only if one, and only one of the conditions is true.

Is there a method of doing that?

You can use the .some method referenced here.

The some() method tests whether at least one element in the array passes the test implemented by the provided function.

// test cases
const str1 = 'hi hello, how do you do?';
const str2 = 'regular string';
const str3 = 'hello there';

// do the test strings contain these terms?
const conditions = ["hello", "hi", "howdy"];

// run the tests against every element in the array
const test1 = conditions.some(el => str1.includes(el));
const test2 = conditions.some(el => str2.includes(el));
// strictly check that contains 1 and only one match
const test3 = conditions.reduce((a,c) => a + str3.includes(c), 0) == 1;

// display results
console.log(`Loose matching, 2 matches "${str1}" => ${test1}`);
console.log(`Loose matching, 0 matches "${str2}" => ${test2}`);
console.log(`Exact matching, 1 matches "${str3}" => ${test3}`);

Also, as a user mentions below, it is also interesting to match "exactly one" appearance like mentioned above (and requested by OP). This can be done similarly counting the intersections with .reduce and checking later that they're equal to 1.

With includes(), no, but you can achieve the same thing with REGEX via test():

var value = /hello|hi|howdy/.test(str);

Or, if the words are coming from a dynamic source:

var words = ['hello', 'hi', 'howdy'];
var value = new RegExp(words.join('|')).test(str);

The REGEX approach is a better idea because it allows you to match the words as actual words, not substrings of other words. You just need the word boundary marker \b, so:

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

That should work even if one, and only one of the conditions is true :

var str = "bonjour le monde vive le javascript";
var arr = ['bonjour','europe', 'c++'];

function contains(target, pattern){
    var value = 0;
    pattern.forEach(function(word){
      value = value + target.includes(word);
    });
    return (value === 1)
}

console.log(contains(str, arr));

You could also do something like this :

const str = "hi, there"

const res = str.includes("hello") || str.includes("hi") || str.includes('howdy');

console.log(res);

Whenever one of your includes return true, value will be true, otherwise, it's going to be false. This works perfectly fine with ES6.

That can be done by using some/every methods of Array and RegEx.

To check whether ALL of words from list(array) are present in the string:

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

To check whether ANY of words from list(array) are present in the string:

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

multiple conditions for JavaScript .includes() method

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