Reputation: 137
I have one JSON Object like this :
var myObject = [
{
"Name" : "app1",
"id" : "1",
"groups" : [
{ "id" : "test1",
"name" : "test group 1",
"desc" : "this is a test group"
},
{ "id" : "test2",
"name" : "test group 2",
"desc" : "this is another test group"
}
]
},
{
"Name" : "app2",
"id" : "2",
"groups" : [
{ "id" : "test3",
"name" : "test group 4",
"desc" : "this is a test group"
},
{ "id" : "test4",
"name" : "test group 4",
"desc" : "this is another test group"
}
]
},
{
"Name" : "app3",
"id" : "3",
"groups" : [
{ "id" : "test5",
"name" : "test group 5",
"desc" : "this is a test group"
},
{ "id" : "test6",
"name" : "test group 6",
"desc" : "this is another test group"
}
]
}
];
I have new value available of "name" for specific "id". How can I replace "name" of specific "id" inside any object ?
And how to count total number of groups among all objects ?
for example : replace name to "test grp45" for id = "test1"
Here is fiddle http://jsfiddle.net/qLTB7/21/
Upvotes: 7
Views: 49095
Reputation: 1
it should be if (object["id"] == value) instead of if (object[x] == value) in 7th line of PitaJ answer, so whole function will look like:
function findAndReplace(object, value, replacevalue) {
for (var x in object) {
if (object.hasOwnProperty(x)) {
if (typeof object[x] == 'object') {
findAndReplace(object[x], value, replacevalue);
}
if (object["id"] == value) {
object["name"] = replacevalue;
// break; // uncomment to stop after first replacement
}
}
}
}
if you leave object[x] - function will replace name also for objects with other keys values set to "test1", for example {"id": "xxx", "name": "test group 1", "desc": "test1"}
Upvotes: 0
Reputation: 15084
The following function will search through an object and all of its child objects/arrays, and replace the key with the new value. It will apply globally, so it won't stop after the first replacement. Uncomment the commented line to make it that way.
function findAndReplace(object, value, replacevalue) {
for (var x in object) {
if (object.hasOwnProperty(x)) {
if (typeof object[x] == 'object') {
findAndReplace(object[x], value, replacevalue);
}
if (object[x] == value) {
object["name"] = replacevalue;
// break; // uncomment to stop after first replacement
}
}
}
}
Working jsfiddle: http://jsfiddle.net/qLTB7/28/
Upvotes: 17
Reputation: 147453
Here's a different approach using Array.prototype.some. It assumes that the Name property in the outer objects should be actually be name (note capitalisation).
function updateNameById(obj, id, value) {
Object.keys(obj).some(function(key) {
if (obj[key].id == id) {
obj[key].name = value;
return true; // Stops looping
}
// Recurse over lower objects
else if (obj[key].groups) {
return updateNameById(obj[key].groups, id, value);
}
})
}
The advantage of some is that it stops as soon as the callback returns true.
Upvotes: 1
Reputation: 5454
Maybe a more succinct sol'n
function changeName(objArray, objId, newName) {
objArray.forEach(function(obj) {
if (obj.id === objId) obj.Name = newName;
});
}
Personally: if this were me, when creating these objects, I would create a new obj and key them by id.
var myApps = {};
myObject.forEach(function(o) {
myApps[o.id] = o;
});
=>
{
"1": {
"Name": "app1",
"id": "1",
"groups": [
{
"id": "test1",
"name": "test group 1",
"desc": "this is a test group"
},
{
"id": "test2",
"name": "test group 2",
"desc": "this is another test group"
}
]
}
}
And then you could just do:
myApps['someId'].name = 'This is my new Name'
Check it out here: http://jsfiddle.net/qLTB7/40/
Upvotes: 1
Reputation: 25892
I think this should work for you:-
var id = 'test1';
var newname = 'test grp45';
var numberOfGruops = 0;
myObject.forEach(function(app){
numberOfGruops += app.groups.length; //Count all groups in this app
app.groups.forEach(function(group){
if(group.id===id)
group.name = newname; // replace the name
});
});
Upvotes: 0
Reputation: 1240
Try this
function findAndReplace(object,keyvalue, name) {
object.map(function (a) {
if (a.groups[0].id == keyvalue) {
a.groups[0].name = name
}
})
}
findAndReplace(myObject,"test1" ,"test grp45");
Upvotes: 1