I am confused about delete operator use on array element.
I found that delete can be used with array elements and it removes the element but doesn't update array length , so basically that element is replaced with undefined.But when i tested it I found that it replaces it with blank.
var test = new Array()
delete test[id];
So i have 3 points regarding this
test[id] id in this is element or index of element, as I read that id can be both. But when I tested it , I found its working only when i pass index.
What if the array has no element and I try to delete by passing any value like delete test[3]?
when i delete an element then it replaces that with undefined or blank?
Don't use delete on array, use splice()
delete will not remove the element actually from array. It'll just set the value to undefined
delete does not alter the length of array
delete Demo
var arr = [1, 2, 3, 4, 5];
delete arr[2];
document.write(arr);
console.log(arr);
document.write('<br />Length: ' + arr.length);
Use splice()
The splice() method changes the content of an array by removing existing elements and/or adding new elements.
splice() will return silently, if you passed an out of bound index to remove.
arr.splice(indexToDelete, 1);
splice() Demo
var arr = [1, 2, 3, 4, 5];
arr.splice(2, 1);
arr.splice(100, 1); // For out of bounds index, it fails silently
document.write(arr);
console.log(arr);
document.write('<br />Length: ' + arr.length);
Related
I have an array with a fixed length of 6.
I want to remove the first element from the array, then shift all the elements left by 1, but the length of the array should remain as 6 (the 6th position of the array can be undefined)
I tried using splice but the array length was reduced to 5 which is not what I want to happen.
What is the best approach to achieve the above?
You can use Array.prototype.slice (which does not mutate the original array unlike splice) to take a copy of the array from the 1st position.
Then use Array.from to get a new array from the copy and mention the length property value of the old array:
const arr = [1, 2, 3, 4, 5, 6]
const newArr = Array.from({length: arr.length, ...arr.slice(1)});
console.log(newArr);
You can shift the first element, then push undefined, as shown below:
const array = [1, 2, 3, 4, 5, 6]
array.shift()
array.push(undefined)
console.log(array)
I have an integer array which looks like this:
var _SingleScannedItemIds = [];
At some point I push some items inside the array itself like following:
_SingleScannedItemIds.push(checkbox_value);
Now my question here is:
How do I know if the array contains anything inside of it?
How do I remove integer numbers from it, based on which parameter?
Can someone help me out? I'm using jQuery...
How do I know if the array contains anything inside of it?
if(_SingleScannedItemIds.length == 0)
How do I remove integer numbers from it, based on which parameter?
_SingleScannedItemIds.splice(pos, 1)
What the above does is it removes 1 element from the position pos.
EDIT
delete _SingleScannedItemIds[pos]
Makes the _SingleScannedItemIds[pos] = undefined and then you can reassign it later with ex. _SingleScannedItemIds[0] = "apple". assuming pos = 0
jQuery Has nothing to do here.
Just use array.length array.splice()
Splice(index, numberOfElementsToRemove)
To get the length of an array:
var array = [0, 1, 2, 4];
console.log(array.length);
Removing items from an array is done by the key:
function remove(array, element) {
const index = array.indexOf(element);
array.splice(index, 1);
}
var array = [0, 1, 2, 4];
remove(array, 1); //removes number 1
console.log(array);
I am not sure what I am missing. Need the new array to return the first and last element of the passed array.
var firstAndLast = function(array){
var newArray = new Array();
newArray [0] = array[0]
newArray [1] = array[3]
return plate;
}
For Example:
var array = ['one', 3, 'cool', 4];
firstAndLast(array); // returns ['one', 4]
You need to access the first and last element, and return that in an array. I'm not sure where plate comes from, but in general your function is 'correct' in the sense that you create a new array and set the first and second elements to the first and last elements of the passed array. You just need to return newArray.
But what if the the 4th element isn't the last one in the array? Then you'd have to make it more general:
var array = ['one', 3, 'cool', 4];
var firstAndLast = function(arr) {
return [arr[0], arr[arr.length - 1]]; //No need for new Array(), and you can just assign items in positions inline
}
console.log(firstAndLast(array));
The above code reduces the function into 3 lines, as it returns an array literal, with the first element as arr[0], the first element of the passed array. The second element is arr[arr.length - 1], which is the last element. Consider this:
[1, 2, 3, 4, 5]
The length currently is 5, and the last element is accessed by arr[4]. Now, just subtract one from the length to get 4, and access the last element. arr[arr.length - 1] in this case would yield 5.
MDN defines the difference below but I'm not understanding the difference. I
created an array and then deleted an element, then tried calling the element and it returned undefined.
var ary=[0,1,2,3,4];
delete ary[3];
ary[3]// returns undefined
ary.length //returns 5
var ary2=[0,1,2,3,4];
ary2[2]=undefined
ary2[2]// returns undefined
ary2.length// returns 5
When you delete an array element, the array length is not affected. This holds even if you delete the last element of the array.
When the delete operator removes an array element, that element is no longer in the array. In the following example, trees[3] is removed with delete.
var trees = ["redwood","bay","cedar","oak","maple"];
delete trees[3];
if (3 in trees) {
// this does not get executed
}
If you want an array element to exist but have an undefined value, use the undefined value instead of the delete operator. In the following example, trees[3] is assigned the value undefined, but the array element still exists:
var trees = ["redwood","bay","cedar","oak","maple"];
trees[3] = undefined;
if (3 in trees) {
// this gets executed
}
Consider this:
var a = [1, 2, 3];
a[1] = undefined;
console.log('1' in a); // true
delete a[1];
console.log('1' in a); // false
console.log(a.length); // 3
console.log(a); // [1, undefined, 3]
If you want to remove in the middle of an array you should use splice:
var a = [1, 2, 3];
a.splice(1, 1);
console.log(a.length); // 2
console.log(a); // [1, 3]
The arguments for splice is:
array.splice(start, deleteCount[, item1[, item2[, ...]]])
See more over at MDN
Since array is nothing but object.. delete removes the '3' key from the object without changing anything else. However setting 3rd element to undefined doesn't remove the key 3 but sets its value to undefined.
I have an Array like
var myArray = new Array;
I have to push some elements to array in such a way that the elements will be replaced with same index.
Example :
myArray.push(1);
myArray.push(2);
myArray.push(3);
so now
myArray[0] = 1
myArray[1] = 2
now when i will push element 3 then
myArray[0] will be replaced with 3 and myArray[1] will be replaced with 1 and the element 2 will be removed.
It will continue according to the number of elements pushed...
Can any body help me with this requirement...
push adds to the end of an array. If you want to add a value to the beginning of an array you can use unshift.
myArray.unshift(3);
You can then use pop to remove the last element:
arr.pop();
DEMO
However, what you might need, given that you need to remove the same number of elements from an array that you add is a function that uses concat and slice instead:
function pusher(arr, add) {
return add.concat(arr).slice(0, arr.length);
}
var arr = [1, 2, 3, 4];
var arr = pusher(arr, [5, 6]); // [5, 6, 1, 2]
DEMO
I think you need something in the lines of:
myArray.unshift(element);
myArray.pop();
Explanation:
unshift: inserts the element on position 0 and moves all other elements one position to the right
pop: removes last element from array