通过对象的嵌套数组布尔属性过滤对象数组

Question

const Person = [
    {
        firstname: "john",
        lastname: "doe",
        items: [
            {
                visible: true,
                foo: "bar"
            },
            {
                visible: false,
                foo: "bar"
            }
        ]
    },
        {
        firstname: "jane",
        lastname: "doe",
        items: [
            {
                visible: false,
                foo: "bar"
            }
        ]
    },
    {
        firstname: "john",
        lastname: "adam",
        items: [
            {
                visible: true,
                foo: "bar"
            },
            {
                visible: false,
                foo: "bar"
            }
        ]
    },
]

In the above array of Person objects, i have an array of items objects that has a property visible . I'm trying to filter Person object based on the visible field: true or false . The filtered Person object's items array of object will only hold true or false as well. How can I achieve this?

I have tried:

I tried:

const filteredPerson = Person.filter(person => {
    person.items.filter(item => item.visible === true)
})

Answer 1

Use Array.some

 const Person = [{firstname: "john",lastname: "doe",items: [{visible: true,foo: "bar"}]},{firstname: "jane",lastname: "doe",items: [{visible: false,foo: "bar"}]},{firstname: "john",lastname: "adam",items: [{visible: true,foo: "bar"}]}]; // Filters all those objects in Person array which has atleast one visible item const result = Person.filter(v => v.items.some(o => o.visible)); console.log(result); 

EDIT (Use nested Array.filter)

 const Person = [{firstname: "john",lastname: "doe",items: [{visible: true,foo: "bar"},{visible: false,foo: "bar"}]},{firstname: "jane",lastname: "doe",items: [{visible: false,foo: "bar"}]},{firstname: "john",lastname: "adam",items: [{visible: true,foo: "bar"},{visible: false,foo: "bar"}]}]; // Filters all those objects in Person array which has atleast one visible item const result = Person.filter(v => { const filteredItems = v.items.filter(o => o.visible); if(filteredItems.length) { // if there is any visible item v.items = filteredItems; return true; } }); console.log(result); 

Answer 2

Single filter() is enough here. Since the visible property in item is at 0 index you can use the that index:

 const Person = [{firstname: "john",lastname: "doe",items: [{visible: true,foo: "bar"}]},{firstname: "jane",lastname: "doe",items: [{visible: false,foo: "bar"}]},{firstname: "john",lastname: "adam",items: [{visible: true,foo: "bar"}]}]; const filteredPerson = Person.filter(person => person.items[0].visible) console.log(filteredPerson); 

If the property index is not known then you have to use some() inside filter() . You can use map() to return the specific property from objects in the array:

 const Person = [{firstname: "john",lastname: "doe",items: [{visible: true,foo: "bar"}]},{firstname: "jane",lastname: "doe",items: [{visible: false,foo: "bar"}]},{firstname: "john",lastname: "adam",items: [{visible: true,foo: "bar"}]}]; const filteredPerson = Person.filter(person => person.items.some(s =>s.visible)).map(p => ({items: p.items})); console.log(filteredPerson);