I'm traversing through an object array:
people = [
{id:1, name:"Bob", available:false},
{id:2, name:"Sally", available:true},
{id:1, name:"Trish", available:false},
]
I want my output to be the names of those available:
["Sally"]
I currently know how to map and extract for a field. How do I add the conditional?
const peopleAvailable = people.map(person => person.value);
Want to do something like this:
const peopleAvailable = people.map(person.available => person.value);
You cannot conditionally map with the .map() function alone, however you can use .filter() to achieve what you require.
Calling filter will return a new array where each item in the new array satisfies your filtering criteria (ie people.available === true).
In the case of your code, you can directly chain filter with your existing call to .map() to obtain the desired result:
const people = [{
id: 1,
name: "Bob",
available: false
},
{
id: 2,
name: "Sally",
available: true
},
{
id: 1,
name: "Trish",
available: false
},
];
const peopleAvailable = people
.filter(people => people.available) /* filter available people */
.map(person => person.name) /* map each person to name */
console.log(peopleAvailable);
=== true in this case.I'll update the answerTry
people.reduce((a,c) => a.concat(c.available ? c.name:[]), [])
people = [
{id:1, name:"Bob", available:false},
{id:2, name:"Sally", available:true},
{id:1, name:"Trish", available:false},
];
let r = people.reduce((a,c) => a.concat(c.available ? c.name:[]), [])
console.log(r);
You can't check the condition in map(). If you put conditions then other elements will be returns as undefined. You can deal with this by applying filter() which allows you to return only those elements which are falls in your condition. So you need to use filter() then use map() on the filtered result.
const array = [{
id: 1,
name: "Bob",
available: false
},
{
id: 2,
name: "Sally",
available: true
},
{
id: 1,
name: "Trish",
available: false
},
];
const withFilter = array
.filter((r) => { return r.available; })
.map((m) => { return m.name; });
console.log(withFilter);
const withOutFilter = array
.map((m) => { if(m.available) { return m.name; } });
console.log(withOutFilter);