How to find if an array contains a specific string in JavaScript/jQuery?

Question

Can someone tell me how to detect if "specialword" appears in an array? Example:

categories: [
    "specialword"
    "word1"
    "word2"
]

Answer 1

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.

Answer 2

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 false

---

Edit another 3 years later

_{Gosh, 6.5 years?!}

The best option for this in modern Javascript is Array.prototype.includes:

var 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.

var categoriesPresent = ['word', 'word', 'specialword', 'word'];
var categoriesNotPresent = ['word', 'word', 'word'];

var foundPresent = categoriesPresent.includes('specialword');
var foundNotPresent = categoriesNotPresent.includes('specialword');

console.log(foundPresent, foundNotPresent); // true false

Answer 3

Here you go:

$.inArray('specialword', arr)

This function returns a positive integer (the array index of the given value), or -1 if the given value was not found in the array.

Live demo: http://jsfiddle.net/simevidas/5Gdfc/

You probably want to use this like so:

if ( $.inArray('specialword', arr) > -1 ) {
    // the value is in the array
}

Answer 4

we can use includes option (which is js built-in function), which will return true if the value is found else it will be false.

if you want the exact index you can use indexOf (which is also js built-in function), which will return the exact index if the value is found else it will return -1.

You can switch .includes with the .some method which returns a boolean. It will exit as soon as a match was found, which is great for performance for huge arrays:

Note: all are case sensitive

var myarr = ["I", "like", "turtles"];

isVal = myarr.includes('like')
index = myarr.indexOf('like')
some = myarr.some(item => item.toLowerCase() == 'like'.toLowerCase())


console.log(isVal)
console.log(index)
console.log(some)

please check this.

Answer 5

You can use a for loop:

var found = false;
for (var i = 0; i < categories.length && !found; i++) {
  if (categories[i] === "specialword") {
    found = true;
    break;
  }
}

Answer 6

With modern javascript's Array methods:

Array.prototype.includes() // introduced in ES7:

• returns boolean

const data = {
  categories: [
    "specialword",
    "word1",
    "word2"
  ]
}

console.log("Array.prototype.includes()")
// Array.prototype.includes()
// returns boolean
console.log(data.categories.includes("specialword"))
console.log(data.categories.includes("non-exist"))

.as-console-wrapper { max-height: 100% !important; top: 0; }

Array.prototype.find() // introduced in ES6:

• returns found element or undefined

const data = {
  categories: [
    "specialword",
    "word1",
    "word2"
  ]
}

console.log("Array.prototype.find()")
// Array.prototype.find()
// returns the element if found
// returns undefined if not found
console.log(data.categories.find(el => el === "specialword") != undefined)
console.log(data.categories.find(el => el === "non-exist") != undefined)

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 7

I don't like $.inArray(..), it's the kind of ugly, jQuery-ish solution that most sane people wouldn't tolerate. Here's a snippet which adds a simple contains(str) method to your arsenal:

$.fn.contains = function (target) {
  var result = null;
  $(this).each(function (index, item) {
    if (item === target) {
      result = item;
    }
  });
  return result ? result : false;
}

Similarly, you could wrap $.inArray in an extension:

$.fn.contains = function (target) {
  return ($.inArray(target, this) > -1);
}