Get all array elements except for first and last
Since there is no data I am taking a basic array to show. Also by this method you will preserve your original array.
var arr = [1,2,3,4,5,6,7];var middle = arr.slice(1, -1);console.log(middle);OR
var arr = [1,2,3,4,5,6,7];var middle = arr.slice(1, arr.length-1);console.log(middle);
To achieve this you can use shift() and pop() to get the first and last elements of the array, respectively. Whatever is left in the array after those operations will be your 'middlepoints'. Try this:
var middlePoints = ['start', 'A', 'B', 'C', 'end'];var origin = middlePoints.shift();var destination = middlePoints.pop();console.log(origin);console.log(middlePoints);console.log(destination);shift and pop affects the current array; Is there a way to get the middlePoints changing the current array?. Like slice, but slice do not affect it
Sure, you can use filter() for that, checking the index of the item in the array:
var input = ['start', 'A', 'B', 'C', 'end'];var output = input.filter((v, i) => i !== 0 && i !== input.length -1);console.log(input); // original value retainedconsole.log(output); // copied via filter
Something like this?
var allPoints = [0, 1, 2, 3, 4, 5], midPoints = []/*start the iteration at index 1 instead of 0 since we want to skip the first point anyway and stop iteration before the final item in the array*/for (var i = 1; i < (allPoints.length - 1); i++) { midPoints.push(allPoints[i]);}console.log(midPoints);