multiple conditions for JavaScript .includes() method
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
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.
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.
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));