There's nothing built-in that will do that for you, you'll have to write a function for it, although it can be just a callback to the some array method.
Two approaches for you:
- Array
somemethod - Regular expression
Array some
The array some method (added in ES5) makes this quite straightforward:
if (substrings.some(function(v) { return str.indexOf(v) >= 0; })) {
// There's at least one
}
Even better with an arrow function and the newish includes method (both ES2015+):
if (substrings.some(v => str.includes(v))) {
// There's at least one
}
Live Example:
const substrings = ["one", "two", "three"];
let str;
// Setup
console.log(`Substrings: ${substrings}`);
// Try it where we expect a match
str = "this has one";
if (substrings.some(v => str.includes(v))) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
// Try it where we DON'T expect a match
str = "this doesn't have any";
if (substrings.some(v => str.includes(v))) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
Regular expression
If you know the strings don't contain any of the characters that are special in regular expressions, then you can cheat a bit, like this:
if (new RegExp(substrings.join("|")).test(string)) {
// At least one match
}
...which creates a regular expression that's a series of alternations for the substrings you're looking for (e.g., one|two) and tests to see if there are matches for any of them, but if any of the substrings contains any characters that are special in regexes (*, [, etc.), you'd have to escape them first and you're better off just doing the boring loop instead. For info about escaping them, see this question's answers.
Live Example:
const substrings = ["one", "two", "three"];
let str;
// Setup
console.log(`Substrings: ${substrings}`);
// Try it where we expect a match
str = "this has one";
if (new RegExp(substrings.join("|")).test(str)) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
// Try it where we DON'T expect a match
str = "this doesn't have any";
if (new RegExp(substrings.join("|")).test(str)) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
Answer from T.J. Crowder on Stack OverflowIs there a reason Array.includes has restrictive typing?
Why includes() works with a string?
Check if an array contains any element of another array in JavaScript - Stack Overflow
How do I check if an array includes a value in JavaScript? - Stack Overflow
Videos
There's nothing built-in that will do that for you, you'll have to write a function for it, although it can be just a callback to the some array method.
Two approaches for you:
- Array
somemethod - Regular expression
Array some
The array some method (added in ES5) makes this quite straightforward:
if (substrings.some(function(v) { return str.indexOf(v) >= 0; })) {
// There's at least one
}
Even better with an arrow function and the newish includes method (both ES2015+):
if (substrings.some(v => str.includes(v))) {
// There's at least one
}
Live Example:
const substrings = ["one", "two", "three"];
let str;
// Setup
console.log(`Substrings: ${substrings}`);
// Try it where we expect a match
str = "this has one";
if (substrings.some(v => str.includes(v))) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
// Try it where we DON'T expect a match
str = "this doesn't have any";
if (substrings.some(v => str.includes(v))) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
Regular expression
If you know the strings don't contain any of the characters that are special in regular expressions, then you can cheat a bit, like this:
if (new RegExp(substrings.join("|")).test(string)) {
// At least one match
}
...which creates a regular expression that's a series of alternations for the substrings you're looking for (e.g., one|two) and tests to see if there are matches for any of them, but if any of the substrings contains any characters that are special in regexes (*, [, etc.), you'd have to escape them first and you're better off just doing the boring loop instead. For info about escaping them, see this question's answers.
Live Example:
const substrings = ["one", "two", "three"];
let str;
// Setup
console.log(`Substrings: ${substrings}`);
// Try it where we expect a match
str = "this has one";
if (new RegExp(substrings.join("|")).test(str)) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
// Try it where we DON'T expect a match
str = "this doesn't have any";
if (new RegExp(substrings.join("|")).test(str)) {
console.log(`Match using "${str}"`);
} else {
console.log(`No match using "${str}"`);
}
One line solution
substringsArray.some(substring=>yourBigString.includes(substring))
Returns true/false if substring exists/doesn't exist
Needs ES6 support
I just stumbled across this: https://www.reddit.com/r/typescript/s/h9MOx2opXV
And it brought back a thought I’ve had a few times.
Is there a good reason Array.includes restricts the search argument to extending the type of the array, given that (to my understanding), any value can be passed to includes and checked against any contents.
Unless I’m missing something it feels like it would be much more useful as a type guard!
Vanilla JS
const found = array1.some(r=> array2.includes(r))
How it works
some(..) checks each element of the array against a test function and returns true if any element of the array passes the test function, otherwise, it returns false. includes(..) both return true if the given argument is present in the array.
vanilla js
/**
* @description determine if an array contains one or more items from another array.
* @param {array} haystack the array to search.
* @param {array} arr the array providing items to check for in the haystack.
* @return {boolean} true|false if haystack contains at least one item from arr.
*/
var findOne = function (haystack, arr) {
return arr.some(function (v) {
return haystack.indexOf(v) >= 0;
});
};
As noted by @loganfsmyth you can shorten it in ES2016 to
/**
* @description determine if an array contains one or more items from another array.
* @param {array} haystack the array to search.
* @param {array} arr the array providing items to check for in the haystack.
* @return {boolean} true|false if haystack contains at least one item from arr.
*/
const findOne = (haystack, arr) => {
return arr.some(v => haystack.includes(v));
};
or simply as arr.some(v => haystack.includes(v));
If you want to determine if the array has all the items from the other array, replace some() to every()
or as arr.every(v => haystack.includes(v));
Modern browsers have Array#includes, which does exactly that and is widely supported by everyone except IE:
console.log(['joe', 'jane', 'mary'].includes('jane')); // true
You can also use Array#indexOf, which is less direct, but doesn't require polyfills for outdated browsers.
console.log(['joe', 'jane', 'mary'].indexOf('jane') >= 0); // true
Many frameworks also offer similar methods:
- jQuery:
$.inArray(value, array, [fromIndex]) - Underscore.js:
_.contains(array, value)(also aliased as_.includeand_.includes) - Dojo Toolkit:
dojo.indexOf(array, value, [fromIndex, findLast]) - Prototype:
array.indexOf(value) - MooTools:
array.indexOf(value) - MochiKit:
findValue(array, value) - MS Ajax:
array.indexOf(value) - Ext:
Ext.Array.contains(array, value) - Lodash:
_.includes(array, value, [from])(is_.containsprior 4.0.0) - Ramda:
R.includes(value, array)
Notice that some frameworks implement this as a function, while others add the function to the array prototype.
Update from 2019: This answer is from 2008 (11 years old!) and is not relevant for modern JS usage. The promised performance improvement was based on a benchmark done in browsers of that time. It might not be relevant to modern JS execution contexts. If you need an easy solution, look for other answers. If you need the best performance, benchmark for yourself in the relevant execution environments.
As others have said, the iteration through the array is probably the best way, but it has been proven that a decreasing while loop is the fastest way to iterate in JavaScript. So you may want to rewrite your code as follows:
function contains(a, obj) {
var i = a.length;
while (i--) {
if (a[i] === obj) {
return true;
}
}
return false;
}
Of course, you may as well extend Array prototype:
Array.prototype.contains = function(obj) {
var i = this.length;
while (i--) {
if (this[i] === obj) {
return true;
}
}
return false;
}
And now you can simply use the following:
alert([1, 2, 3].contains(2)); // => true
alert([1, 2, 3].contains('2')); // => false