Move an element frome an array up or down.

move.js

var move = function(array, element, delta) {
  var index = array.indexOf(element);
  var newIndex = index + delta;
  if (newIndex < 0  || newIndex == array.length) return; //Already at the top or bottom.
  var indexes = [index, newIndex].sort(); //Sort the indixes
  array.splice(indexes[0], 2, array[indexes[1]], array[indexes[0]]); //Replace from lowest index, two elements, reverting the order
};

var moveUp = function(array, element) {
  move(array, element, -1);
};

var moveDown = function(array, element) {
  move(array, element, 1);
};

//Test
var array = [1, 2, 3, 4, 5];
moveUp(array, 4);
moveUp(array, 2);
moveDown(array, 5);

var indexes = [index, newIndex].sort(); //Sort the indixes
Here is the bug, which is based on the fact that default sort acts with indices as strings. So when elements are more than 9, it will place 10 before 9.

You can correct the bug by using a custom sort function that assumes the indices are numeric, modify line 5 to:
var indexes = [index, newIndex].sort((a, b) => a - b); // or for older javascript use: function(a,b) { return a-b; }

var array = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11];
moveUp(array, 11); // failed in original version, passes with correction mentioned above.

Thanks, you helped me a lot!

Thanks soo much, very helpful.

A better alternative to move by index directly
(also fixed the issue when moving with index 10)

move(array, index, delta) {
      //ref: https://gist.github.com/albertein/4496103
      console.log('move', array, index, delta);

      var newIndex = index + delta;
      if (newIndex < 0 || newIndex == array.length) return; //Already at the top or bottom.
      var indexes = [index, newIndex].sort((a, b) => a - b); //Sort the indixes (fixed)
      array.splice(indexes[0], 2, array[indexes[1]], array[indexes[0]]); //Replace from lowest index, two elements, reverting the order
    },