如何根据JavaScript中的值检查对象是否在数组中？

Question

For example, how do I check if ALL the objects in array2 exist in array1 based on their id ?

const array1 = [{ id: 1 }, { id: 2 }, { id: 3 }]
const array2 = [{ id: 1 }, { id: 2 }]

For some reason, I couldn't find the solution on Google.

I thought about this:

const result = array1.every(obj1 => {
  // what do use here? includes? contains?
})

console.log(result)

But I'm kind of stuck in the middle of the code. The most logical solution to me was to use includes . However, includes doesn't seem to take a function: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/includes . So I'm not sure if I can check the objects by id.

Answer 1

includes will only be true if both objects are the same reference in memory, which is (probably) not the case. Instead, I'd create a Set of array1 's id s initially, and then check to see if every array2 id is in the set. This way, you only have to iterate over array1 once, at the very beginning ( Set s have O(1) lookup time):

 const array1 = [{ id: 1 }, { id: 2 }, { id: 3 }]; const array2 = [{ id: 1 }, { id: 2 }]; const idSet = new Set(array1.map(({ id }) => id)); console.log( array2.every(({ id }) => idSet.has(id)) ); 

(Arrays do not have a contains function)

Answer 2

var matching = [];

for (var j = 0; j < array1.length; j++) {
    for (var i = 0; i < array2.length; i++) {
        if (array2[i].id === array1[j].id) {
            matching.push(array2[i].id);       
        }
    }
}

if (array2.length === matching.length) {
    console.log("All elements exist");
} else {
    console.log("One or more of the elements does not exist");
}

Answer 3

Like this, using Array.prototype.every in combination with Array.prototype.some() :

const result = array2.every(o2 => array1.some(o1 => o1.id === o2.id));

This will check if every element in array2 has a matching element in array1 .

---

Here's a complete snippet:

 const array1 = [{ id: 1 }, { id: 2 }, { id: 3 }]; const array2 = [{ id: 1 }, { id: 2 }]; const result = array2.every(o2 => array1.some(o1 => o1.id === o2.id)); console.log(result); 

If array1 is large, or used repeatedly to perform this operation, the Set -based approach proposed in other answers is more optimized.

Answer 4

Similar type of question is already asked. You can refer that also on this link Check if every element in one array is in a second array

 const array1 = [{ id: 1 }, { id: 2 }, { id: 3 }]; const array2 = [{ id: 1 }, { id: 2 }]; let array1ids = array1.map(a => a.id); const result = array2.every(a => array1ids.includes(a.id)); console.log(result); const array3 = [{ id: 1 }, { id: 2 }, { id: 3 }]; const array4 = [{ id: 1 }, { id: 6 }]; let array3ids = array3.map(a => a.id); const result2 = array4.every(a => array3ids.includes(a.id)); console.log(result2); 

Answer 5

For large arrays, I recommend pre-processing the array that's used as the criteria to create a Set so that each check is an O(1) lookup and overall O(n) instead of O(n^2). This code makes a couple assumptions though, namely:

• The criteria is based on a select() function of the object , its index in the array, and a reference to the array .
• The selection used as the criteria returns a primitive and can perform an equality check identical to a Set using a === b || (a !== a && b !== b) a === b || (a !== a && b !== b) (to deal with a and b being NaN )

 const has = select => array => { const set = new Set(array.map(select)) return (object, index, array) => set.has(select(object, index, array)) } const hasId = has(({ id }) => id) const array1 = [{ id: 1 }, { id: 2 }, { id: 3 }] const array2 = [{ id: 1 }, { id: 2 }] const result1 = array1.every(hasId(array2)) const result2 = array2.every(hasId(array1)) console.log(result1) console.log(result2)