Destructuring array of objects in es6
If you want to do with this pure JS then follow this code snippet. It will help you.
let myArray = [
{
"_id": "1",
"subdata": [
{
"subid": "11",
"name": "A"
},
{
"subid": "12",
"name": "B"
}
]
},
{
"_id": "2",
"subdata": [
{
"subid": "12",
"name": "B"
},
{
"subid": "33",
"name": "E"
}
]
}
]
const array = myArray.map(x => x.subdata).flat(1)
const isExist = (key,value, a) => {
return a.find(item => item[key] == value)
}
let a = array.reduce((acc, curr) => {
if(!isExist('subid', curr.subid, acc)) {
acc.push(curr)
}
return acc
}, [])
console.log(a)
I believe what you actually want is
const array = someArray.map(x => x.data)
If you really want three variables (Hint: you shouldn't), you can combine that map
ping with destructuring:
const [array0, array1, array2] = someArray.map(x => x.data)
Whether using destructuring would actually be a simplification is debatable but this is how it can be done:
const [
{ data: array0 },
{ data: array1 },
{ data: array2 }
] = someArray
Live Example:
const someArray = [
{ data: 1 },
{ data: 2 },
{ data: 3 }
];
const [
{ data: array0 },
{ data: array1 },
{ data: array2 }
] = someArray
console.log(array0, array1, array2);
What is happening is that you're first extracting each object from someArray
then destructuring each object by extracting the data
property and renaming it:
// these 2 destructuring steps
const [ obj1, obj2, obj3 ] = someArray // step 1
const { data: array0 } = obj1 // step 2
const { data: array1 } = obj2 // step 2
const { data: array2 } = obj3 // step 2
// written together give
const [
{ data: array0 },
{ data: array1 },
{ data: array2 }
] = someArray
Maybe combine destructuring with mapping for (potentially) more readable code:
const [array0, array1, array2] = someArray.map(item => item.data)
Live Example:
const someArray = [
{ data: 1 },
{ data: 2 },
{ data: 3 }
];
const [array0, array1, array2] = someArray.map(item => item.data)
console.log(array0, array1, array2);