arr1中的每一項都是對象,arr2中的每一項也都是對象數組
const a = [ {_id: 1}, {_id: 2}, {_id: 3}, ]; const b = [ {_id: 2}, {_id: 4}, ]
解決方法一:code
const c = a.filter(x => !b.find(y => y._id === x._id)); // [ { _id: 1 }, { _id: 3} ]
解決方法二:對象
const c = a.filter(x => b.every(y => y._id !== x._id));