cryptpad/www/common/treesome.js

93 lines
2.9 KiB
JavaScript

define([], function () {
var tree = {};
var indexOfNode = tree.indexOfNode = function (el) {
if (!(el && el.parentNode)) {
console.log("No parentNode found!");
throw new Error('No parentNode found!');
}
return Array.prototype.indexOf.call(el.parentNode.childNodes, el);
};
// not being used internally, but is useful externally
tree.contains = function (el, root) {
return el && root && root.contains && root.contains(el);
};
var siblingCount = tree.siblingCount = function (el) {
return el.parentNode.childNodes.length;
};
var childCount = tree.childCount = function (el) {
return el.childNodes.length;
};
/* rightmost and leftmost return the deepest right and left
leaf nodes of a tree
*/
var rightmostNode = tree.rightmostNode = function (el) {
var childNodeCount = childCount(el);
if (!childNodeCount) { // no children
return el; // return the element
} else {
return rightmostNode(el.childNodes[childNodeCount - 1]);
}
};
var leftmostNode = tree.leftmostNode = function (el) {
if (childCount(el)) {
return leftmostNode(el.childNodes[0]);
} else {
return el;
}
};
/* previousNode and nextNode traverse child elements of the dom
in the order in which they appear when selected by a cursor.
in particular, these algorithms traverse text nodes, not just tags
*/
var previousNode = tree.previousNode = function (el, root) {
if (!el || el === root) { return null; }
var i = indexOfNode(el);
if (!el.parentNode) { return null; }
if (i === 0) {
if (root && el.parentNode === root.childNodes[0]) { return null; }
return rightmostNode(previousNode(el.parentNode));
} else {
return rightmostNode(el.parentNode.childNodes[i-1]);
}
};
var nextNode = tree.nextNode = function (el, root) {
if (!el || el === root) { return null; }
var i = indexOfNode(el) + 1, // the index of the next node
l = siblingCount(el);
if (i === l) { // out of bounds
if (el.parentNode === root) { return null; }
return nextNode(el.parentNode, root);
} else {
return leftmostNode(el.parentNode.childNodes[i], root);
}
};
tree.orderOfNodes = function (a, b, root) {
// b might not be supplied
if (!b) { return; }
// a and b might be the same element
if (a === b) { return 0; }
var cur = b;
while (cur) {
cur = previousNode(cur, root);
// if you find 'a' while traversing backwards
// they are in the expected order
if (cur === a) { return 1; }
}
// otherwise
return -1;
};
return tree;
});