A simple map
with object spread syntax will do it.
Instead of object spread, Object.assign
can also be used.
Assign unto a new empty object in order to avoid mutating the existing objects.
var array1 = [
{ name: "lang", value: "English" },
{ name: "age", value: "18" }
];
var array2 = [
{ code: "EN", text: "English language" },
{ code: "DE", value: "German", text: "German language" }
];
var array3 = array1.map((obj, index) => ({
...obj,
...array2[index]
}));
var array3Alternative = array1.map((obj, index) => Object.assign({}, obj, array2[index]));
console.log(array3);
Note that the order of spread patterns and Object.assign
’s arguments matters: latter properties overwrite former properties.
You can also add more properties, e.g. default properties: Object.assign({ code: "XX", value: "Default" }, obj, array2[index])
or ({ code: "XX", value: "Default", ...obj, ...array2[index] })
.
If you want to mutate the objects in array1
, so that the properties from array2
get added into array1
, simply remove the {},
from Object.assign({}, obj, array2[index])
.
Then it would also be advisable to change the map
to a forEach
if you’re not expecting to create a resulting variable like array3
.
If you have an unknown number of arrays like this:
const arrays = [
array1,
array2,
array3,
// …
];
then you can use this approach:
const mergedArray = Array.from({
length: arrays[0].length
}, (_, index) => Object.assign({}, ...arrays.map(({[index]: obj}) => obj))));
See this in action with the following example:
const arrays = [
[
{ a: 3, b: 5 },
{ a: 7, b: 2 },
{ a: 1 }
],
[
{ b: 8, c: 42 },
{ a: 1, b: 12, c: 44 },
{ b: 0 }
],
[
{ d: 14, e: 15 },
{ d: 7 },
{ a: 10 }
]
];
console.log(Array.from({
length: arrays[0].length
}, (_, index) => Object.assign({}, ...arrays.map(({[index]: obj}) => obj))));
Note that this also works with arrays of different lengths.
The resulting value from accessing a non-existent array index is undefined
and spread syntax and Object.assign
take care of this by ignoring it.
Just make sure to start with the longest array, so the result isn’t missing any objects.
This is easier with the snippet where you have an unknown number of arrays (even if it’s always two).
Instead of arrays[0].length
, you just need to use Math.max(...arrays.map(({ length }) => length))
.