I have a scenario where i need to copy the array of Objects(Main array) to another Temp array which should not have object reference basically if i make any modification to Main array it should not reflect in the Temp array so that i will preserve the copy independently.
I have used one of the code snippet from stack overflow this one does partially like if i delete all objects from the Main array the temp array still hold the value but when i do some modifications in main array and click cancel button iam removing all objects from the main array using array.Removeall(); but the modification still exist in Temp array so which means that object having a reference.
clone: function (existingArray) {
var newObj = (existingArray instanceof Array) ? [] : {};
console.debug('newObj value is ' + newObj);
for (i in existingArray) {
console.debug('i value is' + i);
if (i == 'clone') continue;
console.debug('existingArray[i] value ' + existingArray[i]);
if (existingArray[i] && typeof existingArray[i] == "object") {
newObj[i] = this.clone(existingArray[i]);
} else {
console.debug('in else part ' + existingArray[i]);
newObj[i] = existingArray[i];
}
}
return newObj;
}
my object structure is like
iam using knockout framework.
newObjectCreation = function (localIp, RemoteIp, areaId) {
this.localIP = ko.observable(localIp);
this.remoteIP = ko.observable(RemoteIp);
this.areaId = ko.observable(areaId);
};
template.ProtocolArray.push(new newObjectCreation('', '', '')); // to create default row
please help me in this regard. Thanks in advance.
Using the copy() function is another way of copying an array in Python. In this case, a new array object is created from the original array and this type of copy is called deep copy. If any value is modified in the original or copied array, then it does not create any change on another array.
it("creates separate objects in array", () => { const clone = deepClone(people); clone[0]. name = "Jack"; expect(clone[0]. name). toBe("Jack"); expect(people[0].
Let me understand: you don't want just have a new array, but you want to create a new instance for all objects are present in the array itself? So if you modify one of the objects in the temp array, that changes is not propagated to the main array?
If it's the case, it depends by the values you're keeping in the main array. If these objects are simple objects, and they can be serialized in JSON, then the quickest way is:
var tempArray = JSON.parse(JSON.stringify(mainArray));
If you have more complex objects (like instances created by some your own constructors, html nodes, etc) then you need an approach ad hoc.
Edit:
If you don't have any methods on your newObjectCreation
, you could use JSON
, however the constructor won't be the same. Otherwise you have to do the copy manually:
var tempArray = [];
for (var i = 0, item; item = mainArray[i++];) {
tempArray[i] = new newObjectCreation(item.localIP, item.remoteIP, item.areaId);
}
For some other people with the same question. You could also do it this way.
Using the new es6 features you could create a copy of an array (without reference) and a copy of every object without one level of references.
const copy = array.map(object => ({ ...object }))
It's much more functional and idiomatic IMHO
Note: Spread syntax effectively goes one level deep while copying an array. Therefore, it may be unsuitable for copying multidimensional arrays as the following example shows (it's the same with Object.assign() and spread syntax).
More info: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax
So basically if your objects doesn't have objects as properties. This syntax is everything you need. Unfortunately there is not "out of the box" deep clone feature on the spec but you can always use a library if that's what you need
Browser Compatibility Warning: I think it is part of the specification of Ecma now, but some browsers doesn't have full support of spread syntax jet. But using one of the popular transpilers out there you will be fine
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With