Loop through JSON object and check if particular object exists

Loop through JSON object and check if particular object exists - javascript

I have a JSON object :
[{"box":1,"parent":[],"child":[{"boxId":2}]},{"box":2,"parent":[{"boxId":1}],"child":[]}]
I have a requirement where in I would like to check whether my JSON object has particular box; if yes then check if it has particular child.
eg: check if box 1 exists
if yes then
check if it has child
if yes then
check if it has child boxId=2
How do I do that in javascript/ jquery?
This is how I tried:
var DependantArr=myJSON;
var $hasDependancy;
DependantArr.map(function (boxes) {
if (boxes.box == 2) {
if (boxes.child.length != 0) {
boxes.child.map(function (child) {
$hasDependancy = true;
return false;
});
}
}
This doesn't seem to work as even after I return false it still continues to go in loop. I would like to break the loop if i find a match.
Any suggestion?

You need to iterate over all arrays, you need.
var array = [{ "box": 1, "parent": [], "child": [{ "boxId": 2 }] }, { "box": 2, "parent": [{ "boxId": 1 }], "child": [] }];
function check() {
var found = false;
array.some(function (a) {
if (a.box === 1) {
Array.isArray(a.child) && a.child.some(function (b) {
if (b.boxId === 2) {
found = true;
return true;
}
});
return true;
}
});
return found;
}
document.write(check());
Another solution features a more generic approach, with a given object, which acts as a pattern for the needed items.
[
{ condition: { box: 1 }, nextKey: 'child' },
{ condition: { boxId: 2 } }
]
var array = [{ "box": 1, "parent": [], "child": [{ "boxId": 2 }] }, { "box": 2, "parent": [{ "boxId": 1 }], "child": [] }];
function check(array, conditions) {
function c(a, index) {
var el = conditions[index],
k = Object.keys(el.condition),
v = el.condition[k],
found = false;
return Array.isArray(a) &&
a.some(function (b) {
if (b[k] === v) {
found = true;
if (conditions.length > index + 1) {
found = c(b[el.nextKey], index + 1);
}
return true;
}
}) &&
found;
}
return c(array, 0);
}
document.write(check(array, [{ condition: { box: 1 }, nextKey: 'child' }, { condition: { boxId: 2 } }])+'<br>');
document.write(check(array, [{ condition: { box: 2 }, nextKey: 'parent' }, { condition: { boxId: 1 } }]) + '<br>');

Create a function that will call the filter on your array and return it. The returned value would be an array containing the found object(s) which match(es) your condition(s).
Demo Snippet:
(check the console)
var json = [{"box":1,"parent":[],"child":[{"boxId":2}]},{"box":2,"parent":[{"boxId":1}],"child":[]}];
function search(id, childId) {
return json.filter(function(obj) {
if ((obj.box == id) && (obj.child) && (obj.child.length > 0)) {
return obj.child.filter(function(child) {
return (child.boxId == childId);
});
}
});
}
console.log(search('1', '2')[0]);
console.log(search('2', '2')[0]);

You can make use of recursion. call the same function recursively to check if the element you need is really exist in the array or not.
var json = [{"box":1,"parent":[],"child":[{"boxId":2}]},{"box":2,"parent":[{"boxId":1}],"child":[]}];
var found = false;
function validateJson(data, box, boxId){
for(key in data){
if((data[key].constructor === Object || data[key].constructor === Array) && (key !== box && data[key] !== 1 || key !== boxId && data[key] !== 2)){
arguments.callee(data[key]); // <---calls the same function again.
} else {
found = true; // true only in the case of if "box":1 and "boxId" : 2
}
}
return found;
}
var result = validateJson(json, "box", "boxId");
document.body.innerHTML = '<pre> found : '+JSON.stringify(result) + '</pre>';

Try this
data.forEach(function (el) {
Object.keys(el).forEach(function (property) {
if (el[property] === 'your value to check') {
// do whatever you need here
}
});
});

If this is the only case you need to check, you can use this:
var DependantArr = [{"box": 1, "parent": [], "child": [{"boxId": 3}]}, {"box": 2, "parent": [{"boxId": 1}], "child": []}];
var $hasDependancy = DependantArr.some(function(thisBox) {
if ((thisBox.box === 1) && (thisBox.hasOwnProperty('child'))) {
return thisBox.child.filter(function(thisChild) {
return thisChild.boxId === 2;
}).length > 0;
}
});

I think these will help.
for(var i=DependantArr.length;i>=0;i--) {
return DependantArr[i].box == 1 && DependantArr[i].child.length!=0 &&
DependantArr[i].child[0].boxId==2;
}

Related

How can I filter a JSON object in JavaScript?

I've got the following JSON string:
{
"Alarm":{
"Hello":48,
"World":3,
"Orange":1
},
"Rapid":{
"Total":746084,
"Fake":20970,
"Cancel":9985,
"Word": 2343
},
"Flow":{
"Support":746084,
"About":0,
"Learn":0
}
}
Then I load the above string and convert it to json object:
jsonStr = '{"Alarm":{"Hello":48,"World":3,"Orange":1},"Rapid":{"Total":746084,"Fake":20970,"Cancel":9985},"Flow":{"Support":746084,"About":0,"Learn":0}}';
var jsonObj = JSON.parse(jsonStr);
Now, how can I filter this json object by key name?
E.g., if the filter was "ange", the filtered object would be:
{
"Alarm":{
"Orange":1
}
}
If the filter was "flo", the filtered object would become:
{
"Flow":{
"Support":746084,
"About":0,
"Learn":0
}
}
And if the filter was "wor", the result would be:
{
"Alarm":{
"World": 3,
},
"Rapid":{
"Word": 2343
}
}
Is it possible to achieve this filtering using the filter method?

Beside the given solutions, you could use a recursive style to check the keys.
This proposal gives the opportunity to have more nested objects inside and get only the filtered parts.
function filterBy(val) {
function iter(o, r) {
return Object.keys(o).reduce(function (b, k) {
var temp = {};
if (k.toLowerCase().indexOf(val.toLowerCase()) !== -1) {
r[k] = o[k];
return true;
}
if (o[k] !== null && typeof o[k] === 'object' && iter(o[k], temp)) {
r[k] = temp;
return true;
}
return b;
}, false);
}
var result = {};
iter(obj, result);
return result;
}
var obj = { Alarm: { Hello: 48, "World": 3, Orange: 1 }, Rapid: { Total: 746084, Fake: 20970, Cancel: 9985, Word: 2343 }, Flow: { Support: 746084, About: 0, Learn: 0 }, test: { test1: { test2: { world: 42 } } } };
console.log(filterBy('ange'));
console.log(filterBy('flo'));
console.log(filterBy('wor'));
.as-console-wrapper { max-height: 100% !important; top: 0; }

You can create a function using reduce() and Object.keys() that will check key names with indexOf() and return the desired result.
var obj = {
"Alarm": {
"Hello": 48,
"World": 3,
"Orange": 1
},
"Rapid": {
"Total": 746084,
"Fake": 20970,
"Cancel": 9985,
"Word": 2343
},
"Flow": {
"Support": 746084,
"About": 0,
"Learn": 0
}
}
function filterBy(val) {
var result = Object.keys(obj).reduce(function(r, e) {
if (e.toLowerCase().indexOf(val) != -1) {
r[e] = obj[e];
} else {
Object.keys(obj[e]).forEach(function(k) {
if (k.toLowerCase().indexOf(val) != -1) {
var object = {}
object[k] = obj[e][k];
r[e] = object;
}
})
}
return r;
}, {})
return result;
}
console.log(filterBy('ange'))
console.log(filterBy('flo'))
console.log(filterBy('wor'))

With the filter method I think you mean the Array#filter function. This doesn't work for objects.
Anyway, a solution for your input data could look like this:
function filterObjects(objects, filter) {
filter = filter.toLowerCase();
var filtered = {};
var keys = Object.keys(objects);
for (var i = 0; i < keys.length; i++) {
var key = keys[i];
if (objects.hasOwnProperty(key) === true) {
var object = objects[key];
var objectAsString = JSON.stringify(object).toLowerCase();
if (key.toLowerCase().indexOf(filter) > -1 || objectAsString.indexOf(filter) > -1) {
filtered[key] = object;
}
}
}
return filtered;
}

Javascript/Jquery JSON object to array inside object

I see many topics on this site, but every one deal with single Array.
My need is to convert every object with number as key to array.
For exemple,
I have an object like :
{
"parent":{
"0":{
"child":false
},
"1":{
"child":false
},
"4": {
"child":false
}
}
}
And i would like
{
"parent": [
{
"child":false
},
{
"child":false
},
null,
null,
{
"child":false
}
]
}
This is an exemple, my object can be really deep and content many object like this, so i need a generic function.
UPDATE
My try sor far using code of #Nenad Vracar :
function recursiveIteration(object) {
var newob = {};
for (var property in object) {
if (object.hasOwnProperty(property)) {
if (typeof object[property] == "object"){
var result = {};
var keys = Object.keys(object[property]);
if ($.isNumeric(keys[0])) {
console.log("======> "+property+" is table");
for (var i = 0; i <= keys[keys.length - 1]; i++) {
if (keys.indexOf(i.toString()) != -1) {
result[property] = (result[property] || []).concat(object[property][i]);
} else {
result[property] = (result[property] || []).concat(null);
}
}
newob[property] = result;
recursiveIteration(object[property]);
}
newob[property] = object[property];
recursiveIteration(object[property]);
}else{
newob[property] = object[property];
}
}
}
return newob;
}
And the JSFiddle for live try
Thanks you guys !

I think this is what you want:
var data = {
"parent": {
"0": {
"child": false
},
"1": {
"child": false
},
"4": {
"child": false
}
}
};
var convert = function(data) {
// not an object, return value
if (data === null || typeof data !== 'object')
return data;
var indices = Object.keys(data);
// convert children
for (var i = 0; i < indices.length; i++)
data[indices[i]] = convert(data[indices[i]]);
// check if all indices are integers
var isArray = true;
for (var i = 0; i < indices.length; i++) {
if (Math.floor(indices[i]) != indices[i] || !$.isNumeric(indices[i])) {
isArray = false;
break;
}
}
// all are not integers
if (!isArray) {
return data;
}
// all are integers, convert to array
else {
var arr = [];
for (var i = 0, n = Math.max.apply(null, indices); i <= n; i++) {
if (indices.indexOf(i.toString()) === -1)
arr.push(null);
else
arr.push(data[i]);
}
return arr;
}
};
console.log( convert(data) );
Here is a working jsfiddle with the data you provided in the update.

You can do this with Object.keys() and one for loop
var data = {"parent":{"0":{"child":false},"1":{"child":false},"4":{"child":false}}}, result = {}
var keys = Object.keys(data.parent);
for (var i = 0; i <= keys[keys.length - 1]; i++) {
if (keys.indexOf(i.toString()) != -1) {
result.parent = (result.parent || []).concat(data.parent[i]);
} else {
result.parent = (result.parent || []).concat(null);
}
}
console.log(result)

You might achieve this job with a very simple recursive Object method as follows. Any valid nested object (including arrays) within an object structure will be converted into an array, in which the properties are replaced with indices and values are replaced by items.
Object.prototype.valueToItem = function(){
return Object.keys(this).map(e => typeof this[e] === "object" &&
this[e] !== null &&
!Array.isArray(this[e]) ? this[e].valueToItem()
: this[e]);
};
var o = { name: "terrible",
lastname: "godenhorn",
cars: ["red barchetta", "blue stingray"],
age: 52,
child: { name: "horrible",
lastname: "godenhorn",
cars: ["fiat 124", "tata"],
age: 24,
child:{ name: "badluck",
lastname: "godenhorn",
cars: ["lamborghini countach"],
age: 2,
child: null}}},
a = o.valueToItem();
console.log(a);
Ok modified to the OP's conditions but still generic as much as it can be.
Object.prototype.valueToItem = function(){
var keys = Object.keys(this);
return keys.reduce((p,c) => typeof this[c] === "object" &&
this[c] !== null &&
!Array.isArray(this[c]) ? keys.every(k => Number.isInteger(k*1)) ? (p[c] = this[c].valueToItem(),p)
: this[c].valueToItem()
: this
,new Array(~~Math.max(...keys)).fill(null));
};
var o = {
parent: {
0: {
child : false
},
1: {
child : false
},
4: {
child : {
0: {
child : false
},
3: {
child : false
},
5: {
child : false
}
}
}
}
};
a = o.valueToItem();
console.log(JSON.stringify(a,null,4));

Iterate through JSON and change arrays to strings

I have a JSON object that was returned from an XML to js function. This xml converter creates arrays for every entry even when they should be strings. I cannot modify this original function so therefore I would like to take my final json object, iterate through it, detect if a value is an array of length 1 and, if so, change that array to a string.
Original object:
var json = {
user: {
name: ["bob"],
email: ["bob#example.org"]
},
items: [{
name: ["Object 1"]
},{
name: ["Object 2"]
}]
}
Should become:
var json = {
user: {
name: "bob",
email: "bob#example.org"
},
items: [{
name: "Object 1"
},{
name: "Object 2"
}]
}
I have considered the reviver function but a) I would like to avoid going back to a string and b) I am not sure if that would even work as it will probably just feed me each array element individually.

This recursive function seems to work for this problem:
function simplify(obj) {
for (var k in obj) {
if (Object.prototype.toString.call(obj[k]) == '[object Array]' && obj[k].length == 1) {
obj[k] = obj[k][0];
}
else if (typeof obj[k] == 'object') {
obj[k] = simplify(obj[k]);
}
}
return obj;
}
simplify(json);
Demo: http://jsfiddle.net/xkz4W/

Here's a recursive way to do it:
function flattenArrays(data) {
function processItem(item) {
if (Array.isArray(item)) {
if (item.length === 1 && typeof item[0] === "string") {
data[prop] = item[0];
} else {
for (var i = 0; i < item.length; i++) {
processItem(item[i]);
}
}
} else if (typeof item === "object") {
flattenArrays(item);
}
}
for (var prop in data) {
processItem(data[prop]);
}
}
Working demo: http://jsfiddle.net/jfriend00/L5WKs/

Iterate and search values in a object-literal having multiple levels

Here is what I want to do:
I have a tree (javascript object-literal structure) with multiple levels.
I have a value of a particular key of this object.
I want to search for this exact key-value pair in the structure and return the value of another key as an output.
For clarity following is my object literal:
{
"nodeId": 1081,
"appId": 150,
"displayText": "Welcome here",
"Nodes": [
{
"nodeId": 2000,
"appId": 150,
"displayText": "Buy",
"parentNodeId": 1081,
"Nodes": [
{
"nodeId": 2003,
"appId": 150,
"displayText": "tCars",
"parentNodeId": 2000,
"Nodes": [
{
"nodeId": 2006,
"appId": 150,
"displayText": "Diesel",
"parentNodeId": 2003,
"Nodes": [
{
"nodeId": 2008,
"appId": 150,
"displayText": "Price", //This is what I want as return value.
"parentNodeId": 2006,
"Nodes": [],
"nodeCode": "RN_1_1_2_1_3_2_4_1",
"parentCode": "RN_1_1_2_1_3_2",
"jumpToNode": "RN_1_1" //here is the value that I have with me.
}
],
"nodeCode": "RN_1_1_2_1_3_2",
"parentCode": "RN_1_1_2_1"
}
],
"concatWithHeader": false,
"nodeCode": "RN_1_1_2_1",
"parentCode": "RN_1_1"
}
],
"nodeCode": "RN_1_1",
"parentCode": "RN"
}
],
"nodeCode": "RN",
"parentCode": "ROOT_NODE"
}
2. Value that I have with me is "RN_1_1" against jumpToNode
3. I want to search in this object literal and get the value of the key displayText
I searched and tried few things for this but couldnt get the logic to iterate over the inner Nodes objects.
Method I wrote so far:
function getObjects(tree){
var searchkey="RN_1_1";
var displayText = "displayText";
var nodeCode = "nodeCode";
var returnText;
if (tree.hasOwnProperty(nodeCode)) {
var obj = tree[nodeCode];
if(obj == searchkey){
returnText = tree[displayText]; //gives me the return text
break;
}
else{
//here I should iterate over the inner `Nodes` and get the required value.
}
}
}
Please help.
Thanks.

I think you can do something like this which works recursively:
function findProperty(obj, prop, val, propToFetch) {
var answer;
if (obj.hasOwnProperty(prop) && obj[prop] === val) {
return obj[propToFetch];
}
for (var i = 0, len = obj.Nodes.length; i < len; i++) {
answer = findProperty(obj.Nodes[i], prop, val, propToFetch);
if (answer !== null) {return answer;}
}
return null;
}
var result = findProperty(data, "jumpToNode", "RN_1_1", "displayText");
Working demo here: http://jsfiddle.net/jfriend00/EjC5V/

Accordingly to your JSON object you can use this way:
var searchKey="RN_1_1",
displayText = "displayText",
nodeCode = "nodeCode",
returnText,
treeSearch = function (obj, searchKey) {
if (obj[nodeCode] === searchKey) {
returnText = obj[displayText];
} else {
if (obj['Nodes'][0]) {
treeSearch(obj['Nodes'][0], searchKey);
} else {
returnText = null
}
}
};
treeSearch(JSONdata, 'RN_1_1_2_1_3_2');

I have flattened the array using the nodeId to be easier to search through it.
After you flatten the array you can filter it as you wish(i suggest using underscorejs.org)
Here is the live example. The result is displayed in the console.
function flattenNodes (obj, newArr) {
if (obj && obj.Nodes) {
var nodes = obj.Nodes;
delete(obj.Nodes);
newArr[obj.nodeId] = obj;
return flattenNodes(nodes.pop(), newArr);
} else {
return newArr;
}
};
var flattenArr = flattenNodes(arr, new Array());
function findInJumpToNode(find) {
for(key in flattenArr) {
if (flattenArr[key] && flattenArr[key]['jumpToNode']) {
if (flattenArr[key]['jumpToNode'] == find) {
return flattenArr[key];
}
}
}
}
var found = findInJumpToNode('RN_1_1');
console.log(found);

You can use recursion to handle your case.
Check out this sample on jsFiddle.
var nodes= [getNodes()];
alert(getObjects(nodes));
function getObjects(tree){
var searchkey="RN_1_1_2_1_3_2_4_1";
var displayText = "displayText";
var nodeCode = "nodeCode";
var returnText;
if(tree.length > 0)
{
if(tree[0]["nodeCode"] === searchkey)
{
return tree[0][displayText];
}
if(typeof tree[0]["Nodes"] === "undefined")
{
return;
}
return getObjects(tree[0]["Nodes"]);
}
}

How to find a node in a tree with JavaScript

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;
}

Develop Reference

JavaScript is the programming language of the Web.

Loop through JSON object and check if particular object exists - javascript

Try this data.forEach(function (el) { Object.keys(el).forEach(function (property) { if (el[property] === 'your value to check') { // do whatever you need here } }); });

I think these will help. for(var i=DependantArr.length;i>=0;i--) { return DependantArr[i].box == 1 && DependantArr[i].child.length!=0 && DependantArr[i].child[0].boxId==2; }

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