I want to find all objects in items array which is on groupedItems array by 'id' and add to each item isGrouped: true property.
const items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const groupedItems = [
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
so the result should be:
items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2", isGrouped:true },
{ id: 3, name: "item3", isGrouped: true }
];
any ideas ?
You can iterate over the items array using map, creating an isGrouped property which is the result of calling findIndex on the item.id property in the groupedItems id values:
const items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const groupedItems = [
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const result = items.map(item => ({
...item,
isGrouped : groupedItems.findIndex(g => g.id == item.id) >= 0
}
));
console.log(result);Note that I've added isGrouped as false for those objects which are not grouped, rather than omitting the property. It seems that it should be easier to just test a boolean value of a property rather than checking whether the property exists. If you really want to omit the property, you could do something like this:
const items = [
{ id: 1, name: "item1" },
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const groupedItems = [
{ id: 2, name: "item2" },
{ id: 3, name: "item3" }
];
const result = items.map(item =>
groupedItems.find(g => g.id == item.id) ? { ...item, isGrouped : true } : item
);
console.log(result);