Can you search for 2 values at a time in an array in Javascript

Question

I understand I can search for an ( a single item in an array) several different ways. Iterate with for loops, foreach, recursive structures etc, but I am not trying to do that as I am searching for multiple items at a time.

For instance I may have an array of objects with this model:

people: [{name: "howdy", date: 2/2, sex: male}, {name: "jack", date: 3/3, sex: male}]

I want to know if howdy and jack are there at the same time. or maybe I want to know if howdy and jack are both in the array of people. Or maybe I want to know if howdy and Jack are both there by searching through the people array for matches based on birthday and sex.

I know how to do it one at a time but not searching by multiple values. I found this on stackover flow but I don't understand it. Is there a simpler way to search an array to find matches in the array based on multiple keys?

This did not work:

function containsAll(needles, haystack){

    for(var i = 0, len = needles.length; i <len; i++){
        if($.inArray(needles[i], haystack) == -1_ return false
    }
        return true;

}


containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89]); // true

containsAll([34, 78, 89], [78, 67, 99, 56, 89]); // false

containsAll([34, 78, 89], [78, 89]); // false

Answer 1

Your approach:

Move the return true out from the for-loop

function containsAll(needles, haystack) {

  for (var i = 0, len = needles.length; i < len; i++) {
    if ($.inArray(needles[i], haystack) === -1) {
      return false
    }
  }
  
  return true;
}


console.log(containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89])); // true
console.log(containsAll([34, 78, 89], [78, 67, 99, 56, 89])); // false
console.log(containsAll([34, 78, 89], [78, 89])); // false

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>

You can use the functions every and includes:

function containsAll(needles, haystack) {
  return needles.every(function(p) {
    return haystack.includes(p);
  });
}

console.log(containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89])); // true
console.log(containsAll([34, 78, 89], [78, 67, 99, 56, 89])); // false
console.log(containsAll([34, 78, 89], [78, 89]));

Full ES6:

var containsAll = (needles, haystack) => needles.every(p => haystack.includes(p));
  
console.log(containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89])); // true
console.log(containsAll([34, 78, 89], [78, 67, 99, 56, 89])); // false
console.log(containsAll([34, 78, 89], [78, 89]));

To check the both names are within an array like this:

[{name: "howdy", date: 2/2, sex: male}, {name: "jack", date: 3/3, sex: male}]

Follow this approach:

var containsAll = (needles, haystack) => needles.every(p => haystack.findIndex(h => h.name === p) > -1);

console.log(containsAll(['howdy', 'jack'], [{name: "howdy", date: '2/2', g: 'm'}, {name: "jack", date: '3/3', g: 'm'}]));
console.log(containsAll(['howdy', 'ele'], [{name: "howdy", date: '2/2', g: 'm'}, {name: "jack", date: '3/3', g: 'm'}]));

Following your real scenario:

The condition will be h.animal === p.animal

var containsAll = (needles, haystack) => needles.every(p => haystack.findIndex(h => h.animal === p.animal) > -1);

console.log(containsAll([{
  num: 34,
  animal: "chicken",
  city: "chicago"
}], [{
  number: 78,
  name: "chicken",
  num: 34
}, 99, 56, "hello", {
  num: 34,
  animal: "chicken",
  city: "chicago"
}]));

Answer 2

If these arrays are primitives you can use includes:

function containsAll(needles, haystack){
    return needles.every(x => haystack.includes(x));
}

containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89]); // true
containsAll([34, 78, 89], [78, 67, 99, 56, 89]); // false
containsAll([34, 78, 89], [78, 89]); // false

---

Updated:

If you don't know if arrays are primitives then you must check if it is primitive, if don't then check property by property:

function containsAll(needles, haystack){
  return needles.every(x => haystack.find(y => {
    if(typeof y == 'string' || typeof y == 'number') {
      return x == y;
    }
    return Object.keys(y).every(z => x[z] == y[z]);
  }));
}

console.log(containsAll([34, 78, 89], [78, 67, 34, 99, 56, 89]));
//true

console.log(containsAll([34, 78, 89], [78, 67, 99, 56, 89])); 
// false

console.log(containsAll([34, 78, 89], [78, 89])); 
// false

console.log(containsAll([{num: 34, animal: "chicken", city: "chicago"}], [{number: 78, name: "chicken", num: 34}, 99, 56, "hello", {num: 34, animal: "chicken", city: "chicago"}]));
// true

console.log(containsAll([{num: 35, animal: "chicken", city: "chicago"}], [{number: 78, name: "chicken", num: 34}, 99, 56, "hello", {num: 34, animal: "chicken", city: "chicago"}]));
// false