I have a JavaScript code like so:
var myArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]; for (var i = 0, di = 1; i >= 0; i += di) { if (i == myArray.length - 1) { di = -1; } document.writeln(myArray[i]); } I need it to stop right in the middle like 10 and from 10 starts counting down to 0 back.
So far, I've managed to make it work from 0 to 20 and from 20 - 0.
How can I stop it in a middle and start it from there back?
Please help anyone!
Here is an example using a function which accepts the array and the number of items you want to display forwards and backwards:
var myArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]; if(myArray.length === 1){ ShowXElementsForwardsAndBackwards(myArray, 1); } else if(myArray.length === 0) { //Do nothing as there are no elements in array and dividing 0 by 2 would be undefined } else { ShowXElementsForwardsAndBackwards(myArray, (myArray.length / 2)); } function ShowXElementsForwardsAndBackwards(mYarray, numberOfItems){ if (numberOfItems >= mYarray.length) { throw "More Numbers requested than length of array!"; } for(let x = 0; x < numberOfItems; x++){ document.writeln(mYarray[x]); } for(let y = numberOfItems - 1; y >= 0; y--){ document.writeln(mYarray[y]); } } 
ShowXElementsForwardsAndBackwards(myArray, (myArray.length / 2));? Or in the function doesn't keep the numberOfItems parameter and make a var for the middle of the array...
Just divide your array length by 2
var myArray = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20]; for (var i = 0, di = 1; i >= 0; i += di) { if (i == ((myArray.length / 2) -1 )) { di = -1; } document.writeln(myArray[i]); }Could Array.reverse() help you in this matter?
const array = [0,1,3,4,5,6,7,8,9,10,11,12,13,14,15] const getArrayOfAmount = (array, amount) => array.filter((item, index) => index < amount) let arraySection = getArrayOfAmount(array, 10) let reversed = [...arraySection].reverse() console.log(arraySection) console.log(reversed)And then you can "do stuff" with each array with watever array manipulation you desire.
Couldn’t you just check if you’ve made it halfway and then subtract your current spot from the length?
for(i = 0; i <= myArray.length; i++){ if( Math.round(i/myArray.length) == 1 ){ document.writeln( myArray[ myArray.length - i] ); } else { document.writeln( myArray[i] ); } } Unless I’m missing something?
You could move the checking into the condition block of the for loop.
var myArray = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20]; for ( var i = 0, l = (myArray.length >> 1) - 1, di = 1; i === l && (di = -1), i >= 0; i += di ) { document.writeln(myArray[i]); }
If you capture the midpoint ( half the length of the array ), just start working your step in the opposite direction.
const N = 20; let myArray = [...Array(N).keys()]; let midpoint = Math.round(myArray.length/2) for ( let i=1, step=1; i; i+=step) { if (i === midpoint) step *= -1 document.writeln(myArray[i]) }To make things clearer, I've:
i) at 1; this also meant the array has an unused 0 value at 0 index; in other words, myArray[0]==0 that's never showni, which means when i==0 the loop will stop because it is falsydi to step, which is more consistent with other terminologymidpoint uses a Math.round() to ensure it's the highest integer (midpoint) (e.g., 15/2 == 7.5 but you want it to be 8 )midpoint is a variable for performance reasons; calculating the midpoint in the loop body is redundant and less efficient since it only needs to be calculated onceN...) ... but that's easily avoidable