<% // TODO: code cleanup + putting this behind a toggle/attribute const previousNote = (() => { // If we are at the subRoot, there is no previous if (note.noteId === subRoot.note.noteId) return null; const parent = note.getParentNotes()[0]; const children = parent.getVisibleChildNotes(); const index = children.findIndex(n => n.noteId === note.noteId); // If we are the first child, previous goes up a level // this is already protected by the first if statement if (index === 0) return parent; // We are not the first child at this level so previous // should go to the end of the previous tree let candidate = children[index - 1]; while (candidate.hasVisibleChildren()) { const children = candidate.getVisibleChildNotes(); const lastChild = children[children.length - 1]; candidate = lastChild; } return candidate; })(); const nextNote = (() => { // If this currently active note has children, next // should be the first child if (note.hasVisibleChildren()) return note.getVisibleChildNotes()[0]; let parent = note.getParentNotes()[0]; let children = parent.getVisibleChildNotes(); let index = children.findIndex(n => n.noteId === note.noteId); // If we are not the last of the current level, just go // to the next sibling note if (index !== children.length - 1) return children[index + 1]; // We are the last sibling, we need to find the next ancestral note while (index === children.length - 1) { // If we are already at subRoot level, no reason trying to go higher if (parent.noteId === subRoot.note.noteId) return null; const originalParent = parent; parent = parent.getParentNotes()[0]; children = parent.getVisibleChildNotes(); index = children.findIndex(n => n.noteId === originalParent.noteId); } return children[index + 1]; })(); %>