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}"`);
}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
You really don't need jQuery for this.
var myarr = ["I", "like", "turtles"];
var arraycontainsturtles = (myarr.indexOf("turtles") > -1);
Hint: indexOf returns a number, representing the position where the specified searchvalue occurs for the first time, or -1 if it never occurs
or
function arrayContains(needle, arrhaystack)
{
return (arrhaystack.indexOf(needle) > -1);
}
It's worth noting that array.indexOf(..) is not supported in IE < 9, but jQuery's indexOf(...) function will work even for those older versions.
jQuery offers $.inArray:
Note that inArray returns the index of the element found, so 0 indicates the element is the first in the array. -1 indicates the element was not found.
var categoriesPresent = ['word', 'word', 'specialword', 'word'];
var categoriesNotPresent = ['word', 'word', 'word'];
var foundPresent = $.inArray('specialword', categoriesPresent) > -1;
var foundNotPresent = $.inArray('specialword', categoriesNotPresent) > -1;
console.log(foundPresent, foundNotPresent); // true false<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>Edit 3.5 years later
$.inArray is effectively a wrapper for Array.prototype.indexOf in browsers that support it (almost all of them these days), while providing a shim in those that don't. It is essentially equivalent to adding a shim to Array.prototype, which is a more idiomatic/JSish way of doing things. MDN provides such code. These days I would take this option, rather than using the jQuery wrapper.
var categoriesPresent = ['word', 'word', 'specialword', 'word'];
var categoriesNotPresent = ['word', 'word', 'word'];
var foundPresent = categoriesPresent.indexOf('specialword') > -1;
var foundNotPresent = categoriesNotPresent.indexOf('specialword') > -1;
console.log(foundPresent, foundNotPresent); // true falseEdit another 3 years later
Gosh, 6.5 years?!
The best option for this in modern JavaScript is Array.prototype.includes():
const found = categories.includes('specialword');
No comparisons and no confusing -1 results. It does what we want: it returns true or false. For older browsers it's polyfillable using the code at MDN.
const categoriesPresent = ['word', 'word', 'specialword', 'word'];
const categoriesNotPresent = ['word', 'word', 'word'];
const foundPresent = categoriesPresent.includes('specialword');
const foundNotPresent = categoriesNotPresent.includes('specialword');
console.log(foundPresent, foundNotPresent); // true false