Javascript Return Reference To Array Item
Solution 1:
I tried to use
Array.filter()
function, but it returns a new array instead of a reference to the original array. which I can not use to mutate the original array.
It returns a new array, but the array's entries are still references to the same objects. So filter
is just fine for this.
var filteredUsers = users.filter(function(entry) { return entry.id === 2; });
var item = filteredUsers[0];
item.name = "user2 name updated";
console.log(users[1].name) // "user2 name updated"
The array contains references to the objects, not copies of them. When we do item = users[1]
, or use filter
to get a new array containing a subset of the objects, we get a copy of the object reference, and so the variable now refers to the same object, like this:
+-------+ | users | +-------+ | 0 |--------->+---------------+ | | | id: 1 | | | | name: "name1" | +---------------+ | | +---------------+ | filteredUsers | | | +---------------+ | 1 |--------->+---------------+<---------| 0 | | | | id: 2 | +---------------+ | | | name: "name2" | | | | | +------+ | | +---------------+<---------| item | +-------+ +------+
Changing the object changes the object, and those changes are visible regardless of which reference to the object you use to look at its properties.
Now if you had an array of primitives:
var a = [1, 2, 3, 4, 5];
...then of course, you can't use the approach above because they aren't objects. In that case, you'd use indexOf
to find the index:
var index = a.indexOf(3); // Find the first entry === 3
...and then modify the entry
a[index] = 42;
This also applies to strings (provided they're nice normal string primitives, not things you created via new String()
, which there's virtually never a reason to do).
Solution 2:
To do it similarly to the accepted answer, but a bit more succinctly, let us use Array.find, like so:
var item = users.find(u => u.id === 2)
item.name = "user2 name updated"
console.log(users[1].name) // "user2 name updated"
Explanations are very similar to what has been provided in the accepted answer.
Solution 3:
You can use findIndex(), for example:
var users = [{id:1, name:'name1'},{id:2, name:'name2'}]
var idx = users.findIndex(item => item.id === 2);
var u = users[idx]; // the reference of user2
u.name = "name changed."; // same as `users[idx].name = "name changed.";`
Solution 4:
Similarly to @zing-lee's answer, you can remove one more line of code using find()
:
var users = [{id:1, name:'name1'},{id:2, name:'name2'}]
var user = users.find(item => item.id === 2)
user.name = 'name change'
From the docs: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/find
Return value: The value of the first element in the array that satisfies the provided testing function
Because it's an array of Objects (not primitives), the returned value will be a reference to the original object, so modifying a property of the returned object, will update the property of the object that it is referencing.
Good article here explaining values and references in Javasript: https://codeburst.io/explaining-value-vs-reference-in-javascript-647a975e12a0
Solution 5:
Either
users[1].name = "xx"
or if you want to search by id as you tried to do in the question:
for (i = 0; i < users.length; i++) {
if (users[i].id == "2") {
users[i].name = "xxx";
}
}
edit you wanted something more advanced, how about using map: (the idea is that you can just change the elements that match your query and create a new array based on it. you can use filter to filter them out)
[{a: 1, b: 2}, {a:3, b:2}].map(function (elm) {
if (elm.b == 2) {
elm.a += 5;
}
return elm;
});
this returns:
[{a: 6, b: 2}, {a:8, b:2}]
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