Stop an item from being included based on a boolean value

Question

Stop an item from being included based on a boolean value

I need to iterate through an array of IDs called "world". The idea is that if the ID value in world exists in myArray[n].id, then I want to remove the entire element from myArray. If it doesn't exist, then I want to add it to myArray.

world = ["12424126","12461667","12492468","12761163"]

myArray = [
{"id": "12424126"},
{"id": "12761163"},
{"id": "12492468"}
]

For example, if the first element in world[n] ("12424126") matches { "id": "12424126" } in myArray, I should delete the element { "id": "12424126" }

If the first element in world[n] ("12424126") does not exist in myArray:

myArray.push ({ "id": world[n] }); 
}

for (let n = 0; n <= world.length; n++) {

let ID = world[n]; 

finished = false;

if (myArray.find(x => x.id === ID)) {
let index = _.findIndex(myArray, { "id": ID });
if (index > -1) { myArray.splice(index, 1);
finished = true;}
}

if (!finished) // ISSUE: RECORD GETS ADDED REGARDLESS OF FINISHED TRUE/FALSE
{myArray.push ({ id: ID }); // HOW CAN I SOLVE THIS PROBLEM?
}

}

javascript boolean push

Answer 1

Answer №1

This code snippet accomplishes the desired functionality

planet = ["34567890", "34765678", "34563345", "34123456"];

myList = [{ id: "34567890" }, { id: "34123456" }, { id: "34563345" }];
for (i = 0; i < planet.length; i++) {
  ID = planet[i];

  var position = myList.findIndex((element) => element.id == ID);
  if (position > -1) {
    myList.splice(position, 1);
  } else {
    myList.push({ id: ID });
  }
}

Answer 2

This code snippet accomplishes the desired functionality

planet = ["34567890", "34765678", "34563345", "34123456"];

myList = [{ id: "34567890" }, { id: "34123456" }, { id: "34563345" }];
for (i = 0; i < planet.length; i++) {
  ID = planet[i];

  var position = myList.findIndex((element) => element.id == ID);
  if (position > -1) {
    myList.splice(position, 1);
  } else {
    myList.push({ id: ID });
  }
}

Answer 3

Answer №2

The issue lies in the loop causing finished to switch back from true to false in subsequent iterations. To resolve this, you should exit the loop immediately upon setting finished to true.

An alternative solution involves using a Set:

const world = ["12424126","12461667","12492468","12761163"];
let myArray = [{"id": "12424126"},{"id": "12761163"},{"id": "12492468"}];

const set = new Set(myArray.map(({id}) => id).concat(world));
myArray = Array.from(set, id => ({id}));

console.log(myArray);

If you prefer not to create a new array for myArray, and avoid generating new objects for existing ones, then:

const world = ["12424126","12461667","12492468","12761163"];
let myArray = [{"id": "12424126"},{"id": "12761163"},{"id": "12492468"}];

const set = new Set(myArray.map(({id}) => id).concat(world));
myArray.push(...Array.from(set).slice(myArray.length).map(id => ({id})));

console.log(myArray);

However, this approach assumes that myArray did not initially contain duplicate id values.

Answer 4

The issue lies in the loop causing finished to switch back from true to false in subsequent iterations. To resolve this, you should exit the loop immediately upon setting finished to true.

An alternative solution involves using a Set:

const world = ["12424126","12461667","12492468","12761163"];
let myArray = [{"id": "12424126"},{"id": "12761163"},{"id": "12492468"}];

const set = new Set(myArray.map(({id}) => id).concat(world));
myArray = Array.from(set, id => ({id}));

console.log(myArray);

If you prefer not to create a new array for myArray, and avoid generating new objects for existing ones, then:

const world = ["12424126","12461667","12492468","12761163"];
let myArray = [{"id": "12424126"},{"id": "12761163"},{"id": "12492468"}];

const set = new Set(myArray.map(({id}) => id).concat(world));
myArray.push(...Array.from(set).slice(myArray.length).map(id => ({id})));

console.log(myArray);

However, this approach assumes that myArray did not initially contain duplicate id values.

Stop an item from being included based on a boolean value

Answer №1

Answer №2

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