JavaScript is in array

Question

Try this:

if(blockedTile.indexOf("118") != -1)
{  
   // element found
}

As mentioned before, if your browser supports indexOf(), great! If not, you need to pollyfil it or rely on an utility belt like lodash/underscore.

Just wanted to add this newer ES2016 addition (to keep this question updated):

Array.prototype.includes()

if (blockedTile.includes("118")) {
    // found element
}

Best way to do it in 2019 is by using .includes()

[1, 2, 3].includes(2);     // true
[1, 2, 3].includes(4);     // false
[1, 2, 3].includes(1, 2);  // false

First parameter is what you are searching for. Second parameter is the index position in this array at which to begin searching.

If you need to be crossbrowsy here - there are plenty of legacy answers.

function in_array(needle, haystack){
    var found = 0;
    for (var i=0, len=haystack.length;i<len;i++) {
        if (haystack[i] == needle) return i;
            found++;
    }
    return -1;
}
if(in_array("118",array)!= -1){
//is in array
}

Some browsers support Array.indexOf().

If not, you could augment the Array object via its prototype like so...

if (!Array.prototype.indexOf)
{
  Array.prototype.indexOf = function(searchElement /*, fromIndex */)
  {
    "use strict";

    if (this === void 0 || this === null)
      throw new TypeError();

    var t = Object(this);
    var len = t.length >>> 0;
    if (len === 0)
      return -1;

    var n = 0;
    if (arguments.length > 0)
    {
      n = Number(arguments[1]);
      if (n !== n) // shortcut for verifying if it's NaN
        n = 0;
      else if (n !== 0 && n !== (1 / 0) && n !== -(1 / 0))
        n = (n > 0 || -1) * Math.floor(Math.abs(n));
    }

    if (n >= len)
      return -1;

    var k = n >= 0
          ? n
          : Math.max(len - Math.abs(n), 0);

    for (; k < len; k++)
    {
      if (k in t && t[k] === searchElement)
        return k;
    }
    return -1;
  };
}

Source.

Use Underscore.js

It cross-browser compliant and can perform a binary search if your data is sorted.

_.indexOf

_.indexOf(array, value, [isSorted]) Returns the index at which value can be found in the array, or -1 if value is not present in the array. Uses the native indexOf function unless it's missing. If you're working with a large array, and you know that the array is already sorted, pass true for isSorted to use a faster binary search.

Example

//Tell underscore your data is sorted (Binary Search)
if(_.indexOf(['2','3','4','5','6'], '4', true) != -1){
    alert('true');
}else{
    alert('false');   
}

//Unsorted data works to!
if(_.indexOf([2,3,6,9,5], 9) != -1){
    alert('true');
}else{
    alert('false');   
}