I have a JSON input which can go to any number of levels.
Here is the sample
var testJSON = [
{
'name':'USER1',
'id':1,
'child':[],
},{
'name':'USER2',
'id':2,
'child':[{
'name':'USER2_CHILD1',
'id':21,
'child':[]
},{
'name':'USER2_CHILD2',
'id':22,
'child':[{
'name':'USER2_CHILD2_CHILD1',
'id':221,
'child':[]
}]
}],
},{
'name':'USER3',
'id':3,
'child':[{
'name':'USER3_CHILD1',
'id':31,
'child':[]
}],
}];
I want to add a JSON data in child array by finding matching id using the recursive function. For example, if want to add JSON object at id==1; then it was possible by using for loop but what if I want to add JSON object at id==22 or id==221.
I am trying using below code
var body = '';
function scan(obj)
{
var k;
if (obj instanceof Object) {
for (k in obj){
if (obj.hasOwnProperty(k)){
body += 'scanning property ' + k + '<br/>';
scan( obj[k] );
}
}
} else {
body += 'found value : ' + obj + '<br/>';
};
};
scan(testJSON);
document.getElementById('output').innerHTML = body;
You could use an iteration with a check and return the object, if found.
function getObject(array, id) {
var object;
array.some(o => object = o.id === id && o || getObject(o.child || [], id));
return object;
}
var data = [{ name: "USER1", id: 1, child: [] }, { name: "USER2", id: 2, child: [{ name: "USER2_CHILD1", id: 21, child: [] }, { name: "USER2_CHILD2", id: 22, child: [{ name: "USER2_CHILD2_CHILD1", id: 221, child: [] }] }] }, { name: "USER3", id: 3, child: [{ name: "USER3_CHILD1", id: 31, child: [] }] }];
console.log(getObject(data, 1));
console.log(getObject(data, 21));
console.log(getObject(data, 221));
Try this function, you need to parse JSON before
function insertRecord(id,dataToInsert,jsonInput){
let checkIndex = function (arrayElement){
return arrayElement.id === id;
}
let index = jsonInput.findIndex(checkIndex);
if(index != -1) {
if(jsonInput[index].child) {
jsonInput[index].child.push(dataToInsert);
}
else {
jsonInput[index].child = [dataToInsert];
}
}
else {
jsonInput.forEach(function(arrEle, eleIndex){
if(arrEle.child) {
insertRecord(id,dataToInsert,arrEle.child);
}
});
}
}
insertRecord(22,{
'name':'USER1',
'id':33,
'child':[],
},testJSON);
I have a below json, want to get object whose id = 111 , depth may vary depending upon the json.
object = [
{
id= 1,
name : 'a',
childNodes : [ {
id=11,
name:'aa',
childNodes:[{
id: 111,
name:'aaaa',
childNodes:[]
}]
}]
}]
required output { id: 111, name:'aaaa', childNodes:[] }
Looking for fastest algorithm or method. Data would be really huge of more then 35000 nodes and depth upto 20.
Any help would be appreciated.
Here is a recursive function using some:
function findNested(arr, id) {
var res;
return arr.some(o => res = Object(o).id === id ? o
: findNested(o.childNodes, id) ) && res;
}
var object = [{
id: 1,
name : 'a',
childNodes : [ {
id: 11,
name:'aa',
childNodes:[{
id: 111,
name:'aaaa',
childNodes:[]
}]
}]
}];
console.log(findNested(object, 111));
console.log(findNested(object, 9));
You can create recursive function for this using for...in loop.
var object = [{"id":1,"name":"a","childNodes":[{"id":11,"name":"aa","childNodes":[{"id":111,"name":"aaaa","childNodes":[]}]}]},{"id":2,"name":"a","childNodes":[{"id":22,"name":"aa","childNodes":[{"id":123,"name":"aaaa","childNodes":[]}]}]}]
function findById(data, id) {
for(var i in data) {
var result;
if(data.id == id) return data
if(typeof data[i] == 'object' && (result = findById(data[i], id))) return result
}
}
console.log(findById(object, 111))
console.log(findById(object, 22))
I have an object called Suppliers where the id is always unique as follows:
var suppliers = {
Supplier1: {
id: 1,
name: 'Supplier1',
SVGTemplate: 'svg-supplier1'
},
Supplier2: {
id: 2,
name: 'Supplier2',
SVGTemplate: 'svg-supplier2'
},
Supplier3: {
id: 3,
name: 'Supplier3',
SVGTemplate: 'svg-supplier3'
}
}
How can I return the sub-object (e.g return suppliers.Supplier1) when all I know is the id of the sub object? I tried to use .filter, but that only seems to work on arrays:
function findById(source, id) {
return source.filter(function (obj) {
return +obj.id === +id;
})[0];
}
var supplierarray = findById(suppliers, myKnownID);
return supplierarray;
You need to loop over the object using for-in loop, you tried to use .filter which is for array.
So you can change findById definition to below
var suppliers = {
Supplier1: {
id: 1,
name: 'Supplier1',
SVGTemplate: 'svg-supplier1'
},
Supplier2: {
id: 2,
name: 'Supplier2',
SVGTemplate: 'svg-supplier2'
},
Supplier3: {
id: 3,
name: 'Supplier3',
SVGTemplate: 'svg-supplier3'
}
}
// Should return single object not array, as id is unique as OP said
function findById(source, id) {
for(var key in source){
if(source[key].id === id){
return source[key];
}
}
return null;
}
findById(suppliers,3);// {id: 3, name: "Supplier3", SVGTemplate: "svg-supplier3"}
Object.keys(suppliers).map(key => suppliers[key]).find(({id}) => id === someId);
You can search for any key in its child objects.
Object.prototype.findBy = function(propName, propVal){
for( var i in this ){
if( this[i][ propName ] == propVal ){
return this[i];
}
}
return undefined;
};
var suppliers = {
Supplier1: {
id: 1,
name: 'Supplier1',
SVGTemplate: 'svg-supplier1'
},
Supplier2: {
id: 2,
name: 'Supplier2',
SVGTemplate: 'svg-supplier2'
},
Supplier3: {
id: 3,
name: 'Supplier3',
SVGTemplate: 'svg-supplier3'
}
}
console.log(suppliers.findBy("id",1));
Here i've added function inside prototype of root Object.
I have an array of objects similar to the following block of code:
var arr = [
{
text: 'one',
children: [
{
text: 'a',
children: [
{
text: 'something'
}
]
},
{
text: 'b'
},
{
text: 'c'
}
]
},
{
text: 'two'
},
{
text: 'three'
},
{
text: 'four'
}
];
In the above structure, I want to search a string in text property and I need to perform this search over all the children.
For example, if I search for something, the result should be an array of object in the following form:
[
{
children: [
{
children: [
{
text: 'something'
}
]
}
]
}
];
Notice all the text properties that do not match the input string something have been deleted.
I have come up with the following block of code using Array.prototype.filter. However, I can still see extra properties in the result:
function search(arr, str) {
return arr.filter(function(obj) {
if(obj.children && obj.children.length > 0) {
return search(obj.children, str);
}
if(obj.text === str) {
return true;
}
else {
delete text;
return false;
}
});
}
Here is the fiddle link: https://jsfiddle.net/Lbx2dafg/
What am I doing wrong?
I suggest to use Array#forEach, because filter returns an array, which is needed, but not practical for this purpose, because it returns all children with it.
This proposal generates a new array out of the found items, with the wantes item text and children.
The solution works iterative and recursive. It finds all occurences of the search string.
function filter(array, search) {
var result = [];
array.forEach(function (a) {
var temp = [],
o = {},
found = false;
if (a.text === search) {
o.text = a.text;
found = true;
}
if (Array.isArray(a.children)) {
temp = filter(a.children, search);
if (temp.length) {
o.children = temp;
found = true;
}
}
if (found) {
result.push(o);
}
});
return result;
}
var array = [{ text: 'one', children: [{ text: 'a', children: [{ text: 'something' }] }, { text: 'b' }, { text: 'c' }] }, { text: 'two' }, { text: 'three' }, { text: 'four' }];
console.log(filter(array, 'something'));
Your function search returns an array with object from "parent" level, that's why you "still see extra properties in the result".Secondly, this line delete text; doesn't delete an object or object property - it should be delete obj.text;. Here is solution using additional Array.map fuinction:
function search(arr, str) {
return arr.filter(function(obj) {
if (obj.text !== str) {
delete obj.text;
}
if (obj.children && obj.children.length > 0) {
return search(obj.children, str);
}
if (obj.text === str) {
return true;
} else {
delete obj.text;
return false;
}
});
}
var result = search(arr, 'something').map(function(v) { // filtering empty objects
v['children'] = v['children'].filter((obj) => Object.keys(obj).length);
return {'children':v['children'] };
});
console.log(JSON.stringify(result,0,4));
The output:
[
{
"children": [
{
"children": [
{
"text": "something"
}
]
}
]
}
]
https://jsfiddle.net/75nrmL1o/
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;
}