I have these 2 arrays of information. I want to resolve this problem: (Given the name of an user and a required permission, return true if
the user have that permission and false otherwise.)
const groups = [
{
id: 1,
name: 'admins',
permissions: ['CREATE', 'DELETE', 'READ', 'WRITE'],
},
{
id: 2,
name: 'readers',
permissions: ['READ'],
},
{
id: 3,
name: 'writers',
permissions: ['WRITE'],
},
{
id: 4,
name: 'managers',
permissions: ['CREATE', 'DELETE'],
}
];
const users = [
{
name: 'john',
groups: [1],
},
{
name: 'mary',
groups: [2, 3],
},
{
name: 'alice',
groups: [2]
},
{
name: 'bob',
groups: [3],
},
{
name: 'eve',
groups: [2, 4],
},
];
the code I've written so far:
function userHasPermission(userName, requiredPermission) {
//looping through first array
for (i = 0; i < users.length; i++) {
if (users[i].name === userName) {
// looping through the second array
for (j = 0; j < groups.length; j++) {
//here I don't know how to resolve this comparison, because as far as I understand it, the script has to check if userName is a valid ID inside groups array?
if (users[i].groups.includes(groups[j].id) && (groups[j].permissions.includes(requiredPermission))) {
return true;
} else {
return false;
}
}
}
}
};
userHasPermission("mary", "WRITE");
I'm new to Vanilla JavaScript and I'm stuck at this point, any help will be appreciated!
function userHasPermission(userName, requiredPermission) {
//looping through first array
for (i = 0; i < users.length; i++) {
if (users[i].name === userName) {
// looping through the second array
for (j = 0; j < groups.length; j++) {
if (users[i].groups.includes(groups[j].id) && (groups[j].permissions.includes(requiredPermission))) {
return true;
} else {
continue; // don’t return false yet.
}
}
}
}
// only when you’ve checked all possibilities
// still find no match, now you can return definitely
return false;
};
Above is imperative for-loop style based on your code. I personally prefer more functional code style. For your reference:
function userHasPermission(userName, requiredPermission) {
const targetUser = users.find(user => user.name === userName);
if (!targetUser) return false;
const userGroups = groups.filter(g => targetUser.groups.includes(g.id));
const permissions = userGroups.flatMap(g => g.permissions);
return permissions.includes(requiredPermission);
}
function userHasPermission(userName, requiredPermission) {
const user = users.find(u => u.name === userName);
let result = false;
user.groups.forEach(id => {
const group = groups.find(g => g.id === id);
result = group.permissions.includes(requiredPermission);
});
return result;
};
userHasPermission("mary", "WRITE");
I've taken the following sample from a different question. And I am able to identify the object. But I also need to find our the position of that object. For example:
var arr = [{
Id: 1,
Categories: [{
Id: 1
},
{
Id: 2
},
]
},
{
Id: 2,
Categories: [{
Id: 100
},
{
Id: 200
},
]
}
]
If I want to find the object by the Id of the Categories, I can use the following:
var matches = [];
var needle = 100; // what to look for
arr.forEach(function(e) {
matches = matches.concat(e.Categories.filter(function(c) {
return (c.Id === needle);
}));
});
However, I also need to know the position of the object in the array. For example, if we are looking for object with Id = 100, then the above code will find the object, but how do I find that it's the second object in the main array, and the first object in the Categories array?
Thanks!
Well, if every object is unique (only in one of the categories), you can simply iterate over everything.
var arr = [{
Id: 1,
Categories: [{Id: 1},{Id: 2}]
},
{
Id: 2,
Categories: [{Id: 100},{Id: 200}]
}
];
var needle = 100;
var i = 0;
var j = 0;
arr.forEach(function(c) {
c.Categories.forEach(function(e) {
if(e.Id === needle) {
console.log("Entry is in position " + i + " of the categories and in position " + j + " in its category.");
}
j++;
});
j = 0;
i++;
});
function findInArray(needle /*object*/, haystack /*array of object*/){
let out = [];
for(let i = 0; i < haystack.lenght; i++) {
if(haystack[i].property == needle.property) {
out = {pos: i, obj: haystack[i]};
}
}
return out;
}
if you need the position and have to filter over an property of the object you can use a simple for loop. in this sample your result is an array of new object because there can be more mathches than 1 on the value of the property.
i hope it helps
Iterate over the array and set index in object where match found
var categoryGroups = [{
Id : 1,
Categories : [{
Id : 1
}, {
Id : 2
},
]
}, {
Id : 2,
Categories : [{
Id : 100
}, {
Id : 200
},
]
}
]
var filterVal = [];
var needle = 100;
for (var i = 0; i < categoryGroups.length; i++) {
var subCategory = categoryGroups[i]['Categories'];
for (var j = 0; j < subCategory.length; j++) {
if (subCategory[j]['Id'] == findId) {
filterVal.push({
catIndex : i,
subCatIndex : j,
id : needle
});
}
}
}
console.log(filterVal);
Here is solution using reduce:
var arr = [{ Id: 1, Categories: [{ Id: 1 }, { Id: 2 }, ] }, { Id: 2, Categories: [{ Id: 100 }, { Id: 200 }, ] } ]
const findPositions = (id) => arr.reduce((r,c,i) => {
let indx = c.Categories.findIndex(({Id}) => Id == id)
return indx >=0 ? {mainIndex: i, categoryIndex: indx} : r
}, {})
console.log(findPositions(100)) // {mainIndex: 1, categoryIndex: 0}
console.log(findPositions(1)) // {mainIndex: 0, categoryIndex: 0}
console.log(findPositions(200)) // {mainIndex: 1, categoryIndex: 1}
console.log(findPositions(0)) // {}
Beside the given answers with fixt depth searh, you could take an recursive approach by checking the Categories property for nested structures.
function getPath(array, target) {
var path;
array.some(({ Id, Categories = [] }) => {
var temp;
if (Id === target) {
path = [Id];
return true;
}
temp = getPath(Categories, target);
if (temp) {
path = [Id, ...temp];
return true;
}
});
return path;
}
var array = [{ Id: 1, Categories: [{ Id: 1 }, { Id: 2 },] }, { Id: 2, Categories: [{ Id: 100 }, { Id: 200 }] }];
console.log(getPath(array, 100));
.as-console-wrapper { max-height: 100% !important; top: 0; }
I have and object literal that is essentially a tree that does not have a fixed number of levels. How can I go about searching the tree for a particualy node and then return that node when found in an effcient manner in javascript?
Essentially I have a tree like this and would like to find the node with the title 'randomNode_1'
var data = [
{
title: 'topNode',
children: [
{
title: 'node1',
children: [
{
title: 'randomNode_1'
},
{
title: 'node2',
children: [
{
title: 'randomNode_2',
children:[
{
title: 'node2',
children: [
{
title: 'randomNode_3',
}]
}
]
}]
}]
}
]
}];
Basing this answer off of #Ravindra's answer, but with true recursion.
function searchTree(element, matchingTitle){
if(element.title == matchingTitle){
return element;
}else if (element.children != null){
var i;
var result = null;
for(i=0; result == null && i < element.children.length; i++){
result = searchTree(element.children[i], matchingTitle);
}
return result;
}
return null;
}
Then you could call it:
var element = data[0];
var result = searchTree(element, 'randomNode_1');
Here's an iterative solution:
var stack = [], node, ii;
stack.push(root);
while (stack.length > 0) {
node = stack.pop();
if (node.title == 'randomNode_1') {
// Found it!
return node;
} else if (node.children && node.children.length) {
for (ii = 0; ii < node.children.length; ii += 1) {
stack.push(node.children[ii]);
}
}
}
// Didn't find it. Return null.
return null;
Here's an iterative function using the Stack approach, inspired by FishBasketGordo's answer but taking advantage of some ES2015 syntax to shorten things.
Since this question has already been viewed a lot of times, I've decided to update my answer to also provide a function with arguments that makes it more flexible:
function search (tree, value, key = 'id', reverse = false) {
const stack = [ tree[0] ]
while (stack.length) {
const node = stack[reverse ? 'pop' : 'shift']()
if (node[key] === value) return node
node.children && stack.push(...node.children)
}
return null
}
This way, it's now possible to pass the data tree itself, the desired value to search and also the property key which can have the desired value:
search(data, 'randomNode_2', 'title')
Finally, my original answer used Array.pop which lead to matching the last item in case of multiple matches. In fact, something that could be really confusing. Inspired by Superole comment, I've made it use Array.shift now, so the first in first out behavior is the default.
If you really want the old last in first out behavior, I've provided an additional arg reverse:
search(data, 'randomNode_2', 'title', true)
My answer is inspired from FishBasketGordo's iterativ answer. It's a little bit more complex but also much more flexible and you can have more than just one root node.
/**searchs through all arrays of the tree if the for a value from a property
* #param aTree : the tree array
* #param fCompair : This function will receive each node. It's upon you to define which
condition is necessary for the match. It must return true if the condition is matched. Example:
function(oNode){ if(oNode["Name"] === "AA") return true; }
* #param bGreedy? : us true to do not stop after the first match, default is false
* #return an array with references to the nodes for which fCompair was true; In case no node was found an empty array
* will be returned
*/
var _searchTree = function(aTree, fCompair, bGreedy){
var aInnerTree = []; // will contain the inner children
var oNode; // always the current node
var aReturnNodes = []; // the nodes array which will returned
// 1. loop through all root nodes so we don't touch the tree structure
for(keysTree in aTree) {
aInnerTree.push(aTree[keysTree]);
}
while(aInnerTree.length > 0) {
oNode = aInnerTree.pop();
// check current node
if( fCompair(oNode) ){
aReturnNodes.push(oNode);
if(!bGreedy){
return aReturnNodes;
}
} else { // if (node.children && node.children.length) {
// find other objects, 1. check all properties of the node if they are arrays
for(keysNode in oNode){
// true if the property is an array
if(oNode[keysNode] instanceof Array){
// 2. push all array object to aInnerTree to search in those later
for (var i = 0; i < oNode[keysNode].length; i++) {
aInnerTree.push(oNode[keysNode][i]);
}
}
}
}
}
return aReturnNodes; // someone was greedy
}
Finally you can use the function like this:
var foundNodes = _searchTree(data, function(oNode){ if(oNode["title"] === "randomNode_3") return true; }, false);
console.log("Node with title found: ");
console.log(foundNodes[0]);
And if you want to find all nodes with this title you can simply switch the bGreedy parameter:
var foundNodes = _searchTree(data, function(oNode){ if(oNode["title"] === "randomNode_3") return true; }, true);
console.log("NodeS with title found: ");
console.log(foundNodes);
FIND A NODE IN A TREE :
let say we have a tree like
let tree = [{
id: 1,
name: 'parent',
children: [
{
id: 2,
name: 'child_1'
},
{
id: 3,
name: 'child_2',
children: [
{
id: '4',
name: 'child_2_1',
children: []
},
{
id: '5',
name: 'child_2_2',
children: []
}
]
}
]
}];
function findNodeById(tree, id) {
let result = null
if (tree.id === id) {
return tree;
}
if (Array.isArray(tree.children) && tree.children.length > 0) {
tree.children.some((node) => {
result = findNodeById(node, id);
return result;
});
}
return result;}
You have to use recursion.
var currChild = data[0];
function searchTree(currChild, searchString){
if(currChild.title == searchString){
return currChild;
}else if (currChild.children != null){
for(i=0; i < currChild.children.length; i ++){
if (currChild.children[i].title ==searchString){
return currChild.children[i];
}else{
searchTree(currChild.children[i], searchString);
}
}
return null;
}
return null;
}
ES6+:
const deepSearch = (data, value, key = 'title', sub = 'children', tempObj = {}) => {
if (value && data) {
data.find((node) => {
if (node[key] == value) {
tempObj.found = node;
return node;
}
return deepSearch(node[sub], value, key, sub, tempObj);
});
if (tempObj.found) {
return tempObj.found;
}
}
return false;
};
const result = deepSearch(data, 'randomNode_1', 'title', 'children');
This function is universal and does search recursively.
It does not matter, if input tree is object(single root), or array of objects (many root objects). You can configure prop name that holds children array in tree objects.
// Searches items tree for object with specified prop with value
//
// #param {object} tree nodes tree with children items in nodesProp[] table, with one (object) or many (array of objects) roots
// #param {string} propNodes name of prop that holds child nodes array
// #param {string} prop name of searched node's prop
// #param {mixed} value value of searched node's prop
// #returns {object/null} returns first object that match supplied arguments (prop: value) or null if no matching object was found
function searchTree(tree, nodesProp, prop, value) {
var i, f = null; // iterator, found node
if (Array.isArray(tree)) { // if entry object is array objects, check each object
for (i = 0; i < tree.length; i++) {
f = searchTree(tree[i], nodesProp, prop, value);
if (f) { // if found matching object, return it.
return f;
}
}
} else if (typeof tree === 'object') { // standard tree node (one root)
if (tree[prop] !== undefined && tree[prop] === value) {
return tree; // found matching node
}
}
if (tree[nodesProp] !== undefined && tree[nodesProp].length > 0) { // if this is not maching node, search nodes, children (if prop exist and it is not empty)
return searchTree(tree[nodesProp], nodesProp, prop, value);
} else {
return null; // node does not match and it neither have children
}
}
I tested it localy and it works ok, but it somehow won't run on jsfiddle or jsbin...(recurency issues on those sites ??)
run code :
var data = [{
title: 'topNode',
children: [{
title: 'node1',
children: [{
title: 'randomNode_1'
}, {
title: 'node2',
children: [{
title: 'randomNode_2',
children: [{
title: 'node2',
children: [{
title: 'randomNode_3',
}]
}]
}]
}]
}]
}];
var r = searchTree(data, 'children', 'title', 'randomNode_1');
//var r = searchTree(data, 'children', 'title', 'node2'); // check it too
console.log(r);
It works in http://www.pythontutor.com/live.html#mode=edit (paste the code)
no BS version:
const find = (root, title) =>
root.title === title ?
root :
root.children?.reduce((result, n) => result || find(n, title), undefined)
This is basic recursion problem.
window.parser = function(searchParam, data) {
if(data.title != searchParam) {
returnData = window.parser(searchParam, children)
} else {
returnData = data;
}
return returnData;
}
here is a more complex option - it finds the first item in a tree-like node with providing (node, nodeChildrenKey, key/value pairs & optional additional key/value pairs)
const findInTree = (node, childrenKey, key, value, additionalKey?, additionalValue?) => {
let found = null;
if (additionalKey && additionalValue) {
found = node[childrenKey].find(x => x[key] === value && x[additionalKey] === additionalValue);
} else {
found = node[childrenKey].find(x => x[key] === value);
}
if (typeof(found) === 'undefined') {
for (const item of node[childrenKey]) {
if (typeof(found) === 'undefined' && item[childrenKey] && item[childrenKey].length > 0) {
found = findInTree(item, childrenKey, key, value, additionalKey, additionalValue);
}
}
}
return found;
};
export { findInTree };
Hope it helps someone.
A flexible recursive solution that will work for any tree
// predicate: (item) => boolean
// getChildren: (item) => treeNode[]
searchTree(predicate, getChildren, treeNode) {
function search(treeNode) {
if (!treeNode) {
return undefined;
}
for (let treeItem of treeNode) {
if (predicate(treeItem)) {
return treeItem;
}
const foundItem = search(getChildren(treeItem));
if (foundItem) {
return foundItem;
}
}
}
return search(treeNode);
}
find all parents of the element in the tree
let objects = [{
id: 'A',
name: 'ObjA',
children: [
{
id: 'A1',
name: 'ObjA1'
},
{
id: 'A2',
name: 'objA2',
children: [
{
id: 'A2-1',
name: 'objA2-1'
},
{
id: 'A2-2',
name: 'objA2-2'
}
]
}
]
},
{
id: 'B',
name: 'ObjB',
children: [
{
id: 'B1',
name: 'ObjB1'
}
]
}
];
let docs = [
{
object: {
id: 'A',
name: 'docA'
},
typedoc: {
id: 'TD1',
name: 'Typde Doc1'
}
},
{
object: {
id: 'A',
name: 'docA'
},
typedoc: {
id: 'TD2',
name: 'Typde Doc2'
}
},
{
object: {
id: 'A1',
name: 'docA1'
},
typedoc: {
id: 'TDx1',
name: 'Typde Doc x1'
}
},
{
object: {
id: 'A1',
name: 'docA1'
},
typedoc: {
id: 'TDx2',
name: 'Typde Doc x1'
}
},
{
object: {
id: 'A2',
name: 'docA2'
},
typedoc: {
id: 'TDx2',
name: 'Type de Doc x2'
}
},
{
object: {
id: 'A2-1',
name: 'docA2-1'
},
typedoc: {
id: 'TDx2-1',
name: 'Type de Docx2-1'
},
},
{
object: {
id: 'A2-2',
name: 'docA2-2'
},
typedoc: {
id: 'TDx2-2',
name: 'Type de Docx2-2'
},
},
{
object: {
id: 'B',
name: 'docB'
},
typedoc: {
id: 'TD1',
name: 'Typde Doc1'
}
},
{
object: {
id: 'B1',
name: 'docB1'
},
typedoc: {
id: 'TDx1',
name: 'Typde Doc x1'
}
}
];
function buildAllParents(doc, objects) {
for (let o = 0; o < objects.length; o++) {
let allParents = [];
let getAllParents = (o, eleFinded) => {
if (o.id === doc.object.id) {
doc.allParents = allParents;
eleFinded = true;
return { doc, eleFinded };
}
if (o.children) {
allParents.push(o.id);
for (let c = 0; c < o.children.length; c++) {
let { eleFinded, doc } = getAllParents(o.children[c], eleFinded);
if (eleFinded) {
return { eleFinded, doc };
} else {
continue;
}
}
}
return { eleFinded };
};
if (objects[o].id === doc.object.id) {
doc.allParents = [objects[o].id];
return doc;
} else if (objects[o].children) {
allParents.push(objects[o].id);
for (let c = 0; c < objects[o].children.length; c++) {
let eleFinded = null;`enter code here`
let res = getAllParents(objects[o].children[c], eleFinded);
if (res.eleFinded) {
return res.doc;
} else {
continue;
}
}
}
}
}
docs = docs.map(d => buildAllParents(d, objects`enter code here`))
This is an iterative breadth first search. It returns the first node that contains a child of a given name (nodeName) and a given value (nodeValue).
getParentNode(nodeName, nodeValue, rootNode) {
const queue= [ rootNode ]
while (queue.length) {
const node = queue.shift()
if (node[nodeName] === nodeValue) {
return node
} else if (node instanceof Object) {
const children = Object.values(node)
if (children.length) {
queue.push(...children)
}
}
}
return null
}
It would be used like this to solve the original question:
getParentNode('title', 'randomNode_1', data[0])
Enhancement of the code based on "Erick Petrucelli"
Remove the 'reverse' option
Add multi-root support
Add an option to control the visibility of 'children'
Typescript ready
Unit test ready
function searchTree(
tree: Record<string, any>[],
value: unknown,
key = 'value',
withChildren = false,
) {
let result = null;
if (!Array.isArray(tree)) return result;
for (let index = 0; index < tree.length; index += 1) {
const stack = [tree[index]];
while (stack.length) {
const node = stack.shift()!;
if (node[key] === value) {
result = node;
break;
}
if (node.children) {
stack.push(...node.children);
}
}
if (result) break;
}
if (withChildren !== true) {
delete result?.children;
}
return result;
}
And the tests can be found at: https://gist.github.com/aspirantzhang/a369aba7f84f26d57818ddef7d108682
Wrote another one based on my needs
condition is injected.
path of found branch is available
current path could be used in condition statement
could be used to map the tree items to another object
// if predicate returns true, the search is stopped
function traverse2(tree, predicate, path = "") {
if (predicate(tree, path)) return true;
for (const branch of tree.children ?? [])
if (traverse(branch, predicate, `${path ? path + "/" : ""}${branch.name}`))
return true;
}
example
let tree = {
name: "schools",
children: [
{
name: "farzanegan",
children: [
{
name: "classes",
children: [
{ name: "level1", children: [{ name: "A" }, { name: "B" }] },
{ name: "level2", children: [{ name: "C" }, { name: "D" }] },
],
},
],
},
{ name: "dastgheib", children: [{ name: "E" }, { name: "F" }] },
],
};
traverse(tree, (branch, path) => {
console.log("searching ", path);
if (branch.name === "C") {
console.log("found ", branch);
return true;
}
});
output
searching
searching farzanegan
searching farzanegan/classes
searching farzanegan/classes/level1
searching farzanegan/classes/level1/A
searching farzanegan/classes/level1/B
searching farzanegan/classes/level2
searching farzanegan/classes/level2/C
found { name: 'C' }
In 2022 use TypeScript and ES5
Just use basic recreation and built-in array method to loop over the array. Don't use Array.find() because this it will return the wrong node. Use Array.some() instead which allow you to break the loop.
interface iTree {
id: string;
children?: iTree[];
}
function findTreeNode(tree: iTree, id: string) {
let result: iTree | null = null;
if (tree.id === id) {
result = tree;
} else if (tree.children) {
tree.children.some((node) => {
result = findTreeNode(node, id);
return result; // break loop
});
}
return result;
}
const flattenTree = (data: any) => {
return _.reduce(
data,
(acc: any, item: any) => {
acc.push(item);
if (item.children) {
acc = acc.concat(flattenTree(item.children));
delete item.children;
}
return acc;
},
[]
);
};
An Approach to convert the nested tree into an object with depth 0.
We can convert the object in an object like this and can perform search more easily.
The following is working at my end:
function searchTree(data, value) {
if(data.title == value) {
return data;
}
if(data.children && data.children.length > 0) {
for(var i=0; i < data.children.length; i++) {
var node = traverseChildren(data.children[i], value);
if(node != null) {
return node;
}
}
}
return null;
}
In javascript, I have an array of objects that represents an arbitrarily deep list...
data =
[
{ title, depth },
{ title, depth },
{ title, depth },
{ title, depth },
]
...where depth is how deep in the list the element is.
I want to convert this data into html.
For example:
[
{ title: "one", depth : 1 },
{ title: "two", depth : 1 },
{ title: "three", depth : 2 },
{ title: "four", depth : 3 },
{ title: "five", depth : 1 },
]
becomes...
<ul>
<li><p>one</p></li>
<li>
<p>two</p>
<ul>
<li>
<p>three</p>
<ul>
<li><p>four</p></li>
</ul>
</li>
</ul>
</li>
<li><p>five</p></li>
</ul>
What is the simplest way to do this, using jQuery?
Thanks
Here is one way you could do it:
(function($) {
$.listify = function(items) {
var container = $('<div></div>'), root = container;
var depth = 0;
for (var i = 0; i < items.length; ++i) {
var item = items[i];
if (item.depth > depth) {
while (item.depth > depth) {
var ul = $('<ul></ul>').appendTo(container);
var li = $('<li></li>').appendTo(ul);
container = li;
++depth;
}
} else if (item.depth <= depth) {
while (item.depth < depth) {
container = container.parent().parent();
--depth;
}
container = $('<li></li>').insertAfter(container);
}
container.append('<p>' + item.title + '</p>');
}
return root.children();
}
})(jQuery);
I admit, I just wanted to name something listify. Here is a jsFiddle. You would call it like so:
$.listify([
{ title: "one", depth : 1 },
{ title: "two", depth : 1 },
{ title: "three", depth : 2 },
{ title: "four", depth : 3 },
{ title: "five", depth : 1 },
]).appendTo(document.body);
Works with multiple levels and correctly nests the tags:
(function ($) {
$.makelist = function (arr) {
var currentdepth = 1;
var root = $('<ul>');
var el = root;
var idx = 0;
while (idx < arr.length) {
if (arr[idx].depth == currentdepth) {
el.append($('<li>').html($('<p>').text(arr[idx].title)));
idx++;
}
else if (arr[idx].depth > currentdepth) {
newel = $('<ul>');
el.append(newel);
el = newel;
currentdepth++;
}
else {
el = el.parent('ul');
currentdepth--;
}
}
return root;
}
})(jQuery);
$(document).ready(function () {
arr = [
{ title: "one", depth: 1 },
{ title: "two", depth: 1 },
{ title: "three", depth: 2 },
{ title: "four", depth: 3 },
{ title: "five", depth: 1 },
];
$('body').append($.makelist(arr));
});
This one is dynamic:
http://jsfiddle.net/wq8QY/
no matter how many levels you go down it will still be created all the way.
var json = [
{ title: "one", depth : 1 },
{ title: "two", depth : 10 },
{ title: "three", depth : 20 },
{ title: "four", depth : 25 },
{ title: "five", depth : 11 },
]
var dom = $('body') ;
dom.append($('<ul>'))
//console.log(json.length)
for( var i = 0; i < json.length; i++){
//console.log(json[i])
var appendTo = $('body')
for(var j = 1; j <= json[i].depth; j++){
console.log(appendTo, json[i], json[i].depth, '1st')
if(appendTo.children('ul').length <= 0){
console.log('no ul', j)
appendTo.append($('<ul>'))
}
appendTo = $(appendTo.children('ul'))
console.log(appendTo, json[i], '2nd')
}
console.log(appendTo, json[i], 'appending')
appendTo.append($('<li><p>'+json[i].title+'</p></li>'))
}
$.fn.listify = function(list) {
var stack = [this];
$.each(list, function() {
var item = $('<li><p>'+this.title+'</p></li>');
stack[this.depth] = stack[this.depth]
? item.insertAfter(stack[this.depth])
: item.wrap('<ul>').appendTo(stack[this.depth - 1]);
stack.length = this.depth + 1;
});
}