使用some（）根据多个数组过滤一个数组

Question

It's quite possible I am going about this the wrong way, but I have a primary array that i need to filter if any of it's objects values exist in two other arrays. I am trying to use a combination of filter() and some() but what I have right now is not working.

const milestones = <FormArray>this.piForm.get('_milestones');
if (this.piById) {
    milestonesToCreate = milestones.value
        .filter(milestone => !this.piById.milestones.some(item => item.milestoneId === milestone.milestoneId));
    milestonesToDelete = this.piById.milestones
        .filter(milestone => !milestones.value.some(item => item.milestoneId === milestone.milestoneId));
    milestonesToUpdate = milestones.value
        .filter(milestone => milestones.value
            .some(item =>
                item.milestoneId === milestonesToCreate.milestoneId && milestonesToDelete.milestoneId));
}

In the code above milestonesToUpdate should be a the filtered results where the array consists of objects that are not in milestonesToCreate and milestonesToDelete

Hopefully I've explained this well enough.

ADDED SAMPLE MILESTONES ARRAY

milestones = [
  {
    "milestoneId": 0
  }
]

Answer 1

Firstly, it looks like your problem is just a misunderstanding of boolean checks in your final call to some() .

You have put:

item.milestoneId === milestonesToCreate.milestoneId && milestonesToDelete.milestoneId

Which is the same as saying, where item.milestoneId equals milestonesToCreate.milestoneId AND milestonesToDelete.milestoneId exists . I expect that you are just trying to check if the current value exists in both arrays.

Answer 2

it's better to achieve that in single pass:

1. put all elements to elementsToUpdate , copy all elements from your this into elementsToDelete
2. iterate through elementsToUpdate , once some item does not exist in another list, move that element into elementsToCreate
3. if element exists in both, remove it from elementsToDelete .
   4. finally you will get 3 lists you need.

And you can even speed up code more if instead of using arrays you use hash(old good {} ) where id are used as keys. Then check "if element is here" would be as easy as item in elementsToUpdate instead of iterating all the elements each time