JavaScript equivalent of PHP's in_array()

No, it doesn't have one. For this reason most popular libraries come with one in their utility packages. Check out jQuery's inArray and Prototype's Array.indexOf for examples.

jQuery's implementation of it is as simple as you might expect:

function inArray(needle, haystack) {
    var length = haystack.length;
    for(var i = 0; i < length; i++) {
        if(haystack[i] == needle) return true;
    }
    return false;
}

If you are dealing with a sane amount of array elements the above will do the trick nicely.

EDIT: Whoops. I didn't even notice you wanted to see if an array was inside another. According to the PHP documentation this is the expected behavior of PHP's in_array:

$a = array(array('p', 'h'), array('p', 'r'), 'o');

if (in_array(array('p', 'h'), $a)) {
    echo "'ph' was found\n";
}

if (in_array(array('f', 'i'), $a)) {
    echo "'fi' was found\n";
}

if (in_array('o', $a)) {
    echo "'o' was found\n";
}

// Output:
//  'ph' was found
//  'o' was found

The code posted by Chris and Alex does not follow this behavior. Alex's is the official version of Prototype's indexOf, and Chris's is more like PHP's array_intersect. This does what you want:

function arrayCompare(a1, a2) {
    if (a1.length != a2.length) return false;
    var length = a2.length;
    for (var i = 0; i < length; i++) {
        if (a1[i] !== a2[i]) return false;
    }
    return true;
}

function inArray(needle, haystack) {
    var length = haystack.length;
    for(var i = 0; i < length; i++) {
        if(typeof haystack[i] == 'object') {
            if(arrayCompare(haystack[i], needle)) return true;
        } else {
            if(haystack[i] == needle) return true;
        }
    }
    return false;
}

And this my test of the above on it:

var a = [['p','h'],['p','r'],'o'];
if(inArray(['p','h'], a)) {
    alert('ph was found');
}
if(inArray(['f','i'], a)) {
    alert('fi was found');
}
if(inArray('o', a)) {
    alert('o was found');
}  
// Results:
//   alerts 'ph' was found
//   alerts 'o' was found

Note that I intentionally did not extend the Array prototype as it is generally a bad idea to do so.

Array.indexOf was introduced in JavaScript 1.6, but it is not supported in older browsers. Thankfully the chaps over at Mozilla have done all the hard work for you, and provided you with this for compatibility:

if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(elt /*, from*/)
  {
    var len = this.length >>> 0;

    var from = Number(arguments[1]) || 0;
    from = (from < 0)
         ? Math.ceil(from)
         : Math.floor(from);
    if (from < 0)
      from += len;

    for (; from < len; from++)
    {
      if (from in this &&
          this[from] === elt)
        return from;
    }
    return -1;
  };
}

There are even some handy usage snippets for your scripting pleasure.

PHP way:

if (in_array('a', ['a', 'b', 'c'])) {
   // do something if true
}

My solution in JS:

if (['a', 'b', 'c'].includes('a')) {
   // do something if true
}

You can simply use the "includes" function as explained in this lesson on w3schools

it looks like

let myArray = ['Kevin', 'Bob', 'Stuart'];
if( myArray.includes('Kevin'))
console.log('Kevin is here');

There is an equivalent function:

includes()

Look here: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/includes

If you only want to check if a single value is in an array, then Paolo's code will do the job. If you want to check which values are common to both arrays, then you'll want something like this (using Paolo's inArray function):

function arrayIntersect(a, b) {
    var intersection = [];

    for(var i = 0; i < a.length; i++) {
        if(inArray(b, a[i]))
            intersection.push(a[i]);
    }

    return intersection;
}

This wil return an array of values that are in both a and b. (Mathematically, this is an intersection of the two arrays.)

EDIT: See Paolo's Edited Code for the solution to your problem. :)

haystack.find(value => value == needle)

where haystack is an array and needle is an element in array. If element not found will be returned undefined else the same element.