Is there any elegant function to find an array in array? Javascript

actually I want to check if second array has all the keys present in first array I want to check with AND

for example [1,2,3,4].contains([1,4]) must return true where as [1,2,3,4].contains([1,9]) must return false, hope this makes sense, I'm for now using underscores intersection and second array with intersection result it must be same, its still long procedure....

You could use Array.prototype.includes() within a for loop

var arr = [1,2,3,4];

function contains(arr) {
  for (var i = 0, included = true; i < arr.length
       ; included = included && this.includes(arr[i]), i++);
  return included
}

Array.prototype.contains = contains;

console.log(arr.contains([1, 4]), arr.contains([9, 1]), arr.contains([1, 5]))

You can use a combination of every and indexOf

[1, 2].every(function(e) { return [1, 2, 3].indexOf(e) !== -1 });

would evaluate to true and

[1, 5].every(function(e) { return [1, 2, 3].indexOf(e) !== -1 });

would evaluate to false

The syntax would be a lot shorter in ES6 with arrow functions

[1, 2].every(e => [1, 2, 3].indexOf(e) !== -1)

Try this simple example

var arr = [1, 2, 3];
var searchArr = [1,2];
var isSearchArrAvailable = searchArr.filter( function(value){ if ( arr.indexOf( value ) == -1 ) { return true } else { return false; } } ).length == 0;
console.log( "is SearchArr Available " + isSearchArrAvailable );//this will print true

while var searchArr = [1,5] will print false.

You can refactor it into a function

function hasEveryElement(arr, searchArr)
{
   return searchArr.filter( function(value){ if ( arr.indexOf( value ) == -1 ) { return true } else { return false; } } ).length == 0;
}
console.log( hasEveryElement([1,2,3], [1,2]) ) ;
console.log( hasEveryElement([1,2,3], [1,4]) ) ;

You can use every:

function contains(arr, arr2) {
  return arr2.every(function (el) {
    return arr.indexOf(el) > -1;
  });
}

var arr = [1, 2, 3];
contains(arr, [1, 2]); // true
contains(arr, [1, 5]); // false

Alternatively, you could add a method to the prototype if you wanted the [].includes-style syntax, but call it something else - I've used hasAll.

if (!('hasAll' in Array.prototype)) {
  Array.prototype.hasAll = function(arr) {
    return arr.every(function(el) {
      return this.indexOf(el) > -1;
    }, this);
  }

arr.hasAll([1, 2])); // true
arr.hasAll([1, 5])); // false

DEMO

Try this:
var arr = [1,2,3];
var temparr=[1,2];
var st=ArrOfArrSt(arr,temparr); 

function ArrOfArrSt(arr,temparr){
    var stArr= new Array(); 
    if(temparr.length>0)
        for(j in temparr){
                if(arr.indexOf(temparr[j])==-1){
                   stArr.push('0'); 
                }
        }
  return (stArr.length>0)?0:1; //whole array 0=not belong to / 1=belong to  
}