JavaScript is in array

Let's say I have this:

var blockedTile = new Array("118", "67", "190", "43", "135", "520");

There's more array elements but those are just few for readability purposes. Anyways, I could do a "for" loop but it would do 500 loops everytime you click on the map... is there any other way to see if a certain string is in an array?

Answers:

Answer

Try this:

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

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
}
Answer
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
}
Answer

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');   
}
Answer

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.

Answer

Just use for your taste:

var blockedTile = [118, 67, 190, 43, 135, 520];

// includes (js)

if ( blockedTile.includes(118) ){
    console.log('Found with "includes"');
}

// indexOf (js)

if ( blockedTile.indexOf(67) !== -1 ){
    console.log('Found with "indexOf"');
}

// _.indexOf (Underscore library)

if ( _.indexOf(blockedTile, 43, true) ){
    console.log('Found with Underscore library "_.indexOf"');
}

// $.inArray (jQuery library)

if ( $.inArray(190, blockedTile) !== -1 ){
    console.log('Found with jQuery library "$.inArray"');
}
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

Answer
if(array.indexOf("67") != -1) // is in array
Answer

IMHO most compatible with older browsers

Array.prototype.inArray = function( needle ){

    return Array(this).join(",").indexOf(needle) >-1;

}

var foods = ["Cheese","Onion","Pickle","Ham"];
test = foods.inArray("Lemon");
console.log( "Lemon is " + (test ? "" : "not ") + "in the list." );

By turning an Array copy in to a CSV string, you can test the string in older browsers.

Answer

Depending on the version of JavaScript you have available, you can use indexOf:

Returns the first index at which a given element can be found in the array, or -1 if it is not present.

Or some:

Tests whether some element in the array passes the test implemented by the provided function.

But, if you're doing this sort of existence check a lot you'd be better of using an Object to store your strings (or perhaps an object as well as the Array depending on what you're doing with your data).

Answer

I'd use a different data structure, since array seem to be not the best solution.

Instead of array, use an object as a hash-table, like so:

(posted also in jsbin)

var arr = ["x", "y", "z"];
var map = {};
for (var k=0; k < arr.length; ++k) {
  map[arr[k]] = true;
}

function is_in_map(key) {
  try {
    return map[key] === true;
  } catch (e) {
    return false;
  }
}


function print_check(key) {
  console.log(key + " exists? - " + (is_in_map(key) ? "yes" : "no"));
}

print_check("x");
print_check("a");

Console output:

x exists? - yes
a exists? - no

That's a straight-forward solution. If you're more into an object oriented approach, then search Google for "js hashtable".

Answer

Already answered above but wanted to share.

Will not work in IE though. thanks for mentioning it @Mahmoud

var array1 = [1, 2, 3];

console.log(array1.includes(2));
// expected output: true

var pets = ['cat', 'dog', 'bat'];

console.log(pets.includes('cat'));
// expected output: true

console.log(pets.includes('at'));
// expected output: false

got some reference Here. they also have Polyfill for above.

Answer

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.

Answer

Why don't you use Array.filter?

var array = ['x','y','z'];
array.filter(function(item,index,array){return(item==YOURVAL)}).

Just copy that into your code, and here you go:

Array.prototype.inArray = function (searchedVal) {
return this.filter(function(item,index,array){return(item==searchedVal)}).length==true
}
Answer

in array example,Its same in php (in_array)

 var ur_fit = ["slim_fit", "tailored", "comfort"];
 var ur_length = ["length_short", "length_regular", "length_high"];
    if(ur_fit.indexOf(data_this)!=-1){
        alert("Value is avail in ur_fit array");
    }
    else if(ur_length.indexOf(data_this)!=-1){      
         alert("value is avail in ur_legth array"); 

    }
Answer
var myArray = [2,5,6,7,9,6];
myArray.includes(2) // is true
myArray.includes(14) // is false
Answer

You can try below code. Check http://api.jquery.com/jquery.grep/

var blockedTile = new Array("118", "67", "190", "43", "135", "520");
var searchNumber = "11878";
arr = jQuery.grep(blockedTile, function( i ) {
  return i === searchNumber;
});
if(arr.length){ console.log('Present'); }else{ console.log('Not Present'); }

check arr.length if it's more than 0 means string is present else it's not present.

Answer

I think the simplest way is that :

(118 in blockedTile); //is true 

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