I have an array which contains nested arrays that are dynamically created, it looks like this:
[['1', '2'],['1','3', '4'],['1', '3']]
I am trying to implement AND logic by getting only duplicate values from these arrays. My expected output here would be ['1'] since all nested arrays must contain the same value.
// [['1', '2'],['1','3', '4'],['1', '3']]
const arrays = [...new Set(response)].filter(newSet => newSet.length > 0);
const builder = []; // array of all the id's no longer needed
// [[],[],[]]
arrays.forEach(arr => {
// []
arr.forEach(value => {
// [[], [], []]
const found = arrays.filter(a => a.find(v => v === value));
if (found.length === 0)
builder.push(value);
});
});
console.log(builder); // empty []
This gives me an empty array because of the filter(). How could I return an array of values that all (3 in this case but could be more) arrays contain?
Expected output with the above input would be ["1"]. Any help appreciated.
>Solution :
from what I understand you need the common elements from all array
let response1 = [['1', '2'],['1','3', '4'],['1', '3']]
let response2 = [['1', '2', '3'],['1','3', '4'],['1', '3']]
const getCommon = res => [...new Set(res.flat())].filter(a => res.every(c => c.includes(a)));
console.log(getCommon(response1))
console.log(getCommon(response2))