I have a requirement to traverse the model tree and at each node perform some business logic, I was wondering if there are any helper methods that I could use that will help with moving through the nodes. Such as breadth first search or depth first search or even simple methods for getting a nodes children and parents?
Also methods for finding leaf nodes would be very helpful.
(Also if possible it would be great if these where provided in javascript rather than ES6, I know Phillipe Leefsma has a blog article that has a piece of code for finding leaf nodes but it is in ES6 and I can't seem to get it converted back to Javascript)
Any help is greatly appreciated.
The method of getting all the children given an instance tree and a node is
instanceTree.enumNodeChildren(node, callback) where all the children dbids are in the callback.
To traverse through the tree, first get the instancee tree and the root id:
var instanceTree = viewer.model.getData().instanceTree;
var rootId = this.rootId = instanceTree.getRootId();
Then the best way is using breadth-first search:
function getAlldbIds (rootId) {
var alldbId = [];
if (!rootId) {
return alldbId;
}
var queue = [];
queue.push(rootId);
while (queue.length > 0) {
var node = queue.shift();
alldbId.push(node);
instanceTree.enumNodeChildren(node, function(childrenIds) {
queue.push(childrenIds);
});
}
return alldbId;
}
You can modify your method to optimize for performance.
Read more about it here: https://forge.autodesk.com/cloud_and_mobile/2015/12/select-all-elements-in-the-viewer-with-view-and-data-api-with-javascript.html
Related
I want to duplicate my GLTF models with different positions/colors dynamically, to do so I have done:
const L_4_G = new Object3D();
...
const multiLoad_4 = (result, position) => {
const model = result.scene.children[0];
model.position.copy(position);
model.scale.set(0.05, 0.05, 0.05);
//
L_4_G.add(model.clone())
scene.add(model);
};
...
function duplicateModel4() {
L_4_G.translateX(-1.2)
L_4_G.translateY(0.0)//0.48
L_4_G.translateZ(1.2)
L_4_G.rotateY(Math.PI / 2);
scene.add(L_4_G);
}
I didn't find out how can I change the Object3D color from the documentation, can you please tell me how can I do that? thanks in advance.
Here is the full code that I'm using, and here are the models
Update
I have seen this solution, to store a set of colors in the object's userData and choose the color later:
L_2_G.userData.colors = {green : #00FF00, red : ..., ...}
L_2_G.children[0].material.color(userData.colors["green"])
But I'm getting an error that children[0] undefined, but I can see that this object has a child and a material, and color via the console: console.log(L_2_G.children), console.log(L_2_G.children.length)--> 0
Also I have tried getObjectByName as explained here:
scene.getObjectByName(name).children[0].material.color.set(color);
which also reslts: children[0] is undefined, scene.getObjectByName(name).children.length is 0.
THREE.Object3D is a base class for anything that can go in a scene graph, including lights, cameras, and empty objects. Not all Object3D instances have geometry or materials. You may be looking for the THREE.Mesh subclass which does have materials and colors.
In general, code like getObjectByName(...) and model = result.scene.children[0] is very content-specific. The file might contain many nested objects, and .children[0] just grabs the first part. It's usually best to traverse the scene graph instead, looking for the objects you want to modify (e.g. looking for all Meshes, or Meshes with a particular name).
const model = result.scene;
model.traverse((object) => {
if (object.isMesh) {
object.material.color.setHex( 0x404040 );
}
});
Then you can either add the entire group to your scene (scene.add(model)), or just add parts of it. Keep in mind that adding meshes to a new parent removes them from their previous parent, and you shouldn't do that while traversing the previous parent. Instead you can make a list of meshes, and add them in a second step:
const meshes = [];
result.scene.traverse((object) => {
if (object.isMesh) {
meshes.push(object);
}
});
for (const mesh of meshes) {
scene.add(mesh);
}
Finally, the position of an object is inherited from its parents. By removing the object from its original parents you might change its position in the scene. If you are planing to assign a new position to the object anyway, that is fine.
I just wanted to learn Revit API and create a simple wall using ExternalCommand. But I cannot figure it out...
I think my problem is here:
var symbolId = document.GetDefaultFamilyTypeId(new ElementId(BuiltInCategory.OST_Walls));
When I debug it symbolId always -1.
Can you help me what is wrong with this code snippet?
public Autodesk.Revit.UI.Result Execute(
Autodesk.Revit.UI.ExternalCommandData command_data,
ref string message,
Autodesk.Revit.DB.ElementSet elements)
{
var document = command_data.Application.ActiveUIDocument.Document;
var level_id = new ElementId(1526);
// create line
XYZ point_a = new XYZ(-10, 0, 0);
XYZ point_b = new XYZ(10, 10, 10);
Line line = Line.CreateBound(point_a, point_b);
using (var transaction = new Transaction(doc))
{
transaction.Start("create walls");
Wall wall = Wall.Create(doc, line, level_id, false);
var position = new XYZ(0, 0, 0);
var symbolId = document.GetDefaultFamilyTypeId(new ElementId(BuiltInCategory.OST_Walls));
if (symbolId == ElementId.InvalidElementId) {
transaction.RollBack();
return Result.Failed;
}
var symbol = document.GetElement(symbolId) as FamilySymbol;
var level = (Level)document.GetElement(wall.LevelId);
document.Create.NewFamilyInstance(position, symbol, wall, level, StructuralType.NonStructural);
transaction.Commit();
}
return Result.Succeeded;
}
Work through the Revit API getting started material and all will be explained. That will save you and others many further questions and answers.
To address this specific question anyway, GetDefaultFamilyTypeId presumably does not do what you expect it to for wall elements. In the GetDefaultFamilyTypeId method API documentation, it is used for structural columns, a standard loadable family hosted by individual RFA files. Walls are built-in system families and behave differently. Maybe GetDefaultFamilyTypeId only works for non-system families.
To retrieve an arbitrary (not default) wall type, use a filtered element collector to retrieve all WallType elements and pick the first one you find.
Here is a code snippet that picks the first one with a specific name, from The Building Coder discussion on Creating Face Wall and Mass Floor
:
WallType wType = new FilteredElementCollector( doc )
.OfClass( typeof( WallType ) )
.Cast<WallType>().FirstOrDefault( q
=> q.Name == "Generic - 6\" Masonry" );
I am looking for a way to use Javascript to Query Taxonomy.js, to get a Term based on Term Name (I don't have ID available on page).
Only option that I am able to find is to retrieve all terms in the TermSet, loop through each term to match the name.
This works, but is causing performance issue. I am looking for a way to get the term directly, without looping through all.
I was finally able to cobble this together.
var context = SP.ClientContext.get_current();
var session = SP.Taxonomy.TaxonomySession.getTaxonomySession(context);
var termStore = session.getDefaultSiteCollectionTermStore();
var parentTermId = 'd89595cf-7d0d-4f19-8e14-8b8b05efb7de'; // Some parent term
var parentTerm = termStore.getTerm(parentTermId);
// arguments are termLabel, language code, defaultLabelOnly, matching option, max num results, trim unavailable
var terms = parentTerm.getTerms(series,1033,true,SP.Taxonomy.StringMatchOption.exactMatch,1,true);
context.load(terms);
context.executeQueryAsync(
function(){
//print child Terms
for(var i = 0; i < terms.get_count();i++){
var term = terms.getItemAtIndex(i);
console.log(term.get_name());
console.log(term.get_description());
}
},
function(sender,args){
console.log(args.get_message());
});
If you know one parent term's guid you can use that and then get a specific term below that. In our case the parent term only has one level of children, so I haven't checked if it searches children of children.
There is also the TermSet.getTerms() method which takes a labelMatchingInformation object as its argument. Here is the documentation and a blog about it. I couldn't get the lmi to work, but in the blog he seems to be apply it directly to a taxonomysession and I think it is supposed to be applied to a termset, so maybe that is the difference.
I can't seem to figure out how to search in a dijit.Tree, using a ItemFileWriteStore and a TreeStoreModel. Everything is declarative, I am using Dojo 1.7.1, here is what I have so far :
<input type="text" dojoType="dijit.form.TextBox" name="search_fruit" id="search_fruit" onclick="search_fruit();">
<!-- store -->
<div data-dojo-id="fruitsStore" data-dojo-type="dojo.data.ItemFileWriteStore" clearOnClose="true" urlPreventCache="true" data-dojo-props='url:"fruits_store.php"'></div>
<!-- model -->
<div data-dojo-id="fruitsModel" data-dojo-type="dijit.tree.TreeStoreModel" data-dojo-props="store:fruitsStore, query:{}"></div>
<!-- tree -->
<div id="fruitsTree" data-dojo-type="dijit.Tree"
data-dojo-props='"class":"container",
model:fruitsModel,
dndController:"dijit.tree.dndSource",
betweenThreshold:5,
persist:true'>
</div>
The json returned by fruits_store.php is like this :
{"identifier":"id",
"label":"name",
"items":[{"id":"OYAHQIBVbeORMfBNZXFGOHPdaRMNUdWEDRPASHSVDBSKALKIcBZQ","name":"Fruits","children":[{"id":"bSKSVDdRMRfEFNccfTZbWHSACWbLJZMTNHDVVcYGcTBDcIdKIfYQ","name":"Banana"},{"id":"JYDeLNIGPDBRMcfSTMeERZZEUUIOMNEYYcNCaCQbCMIWOMQdMEZA","name":"Citrus","children":[{"id":"KdDUfEDaKOQMFNJaYbSbAcAPFBBdLALFMIPTFaYSeCaDOFaEPbJQ","name":"Orange"},{"id":"SDWbXWbTWKNJDIfdAdJbbbRWcLZFJHdEWASYDCeFOZYdcZUXJEUQ","name":"Lemon"}]},{"id":"fUdQTEZaIeBIWCHMeBZbPdEWWIQBFbVDbNFfJXNILYeBLbWUFYeQ","name":"Common ","children":[{"id":"MBeIUKReBHbFWPDFACFGWPePcNANPVdQLBBXYaTPRXXcTYRTJLDQ","name":"Apple"}]}]}]}
Using a grid instead of a tree, my search_fruit() function would look like this :
function search_fruit() {
var grid = dijit.byId('grid_fruits');
grid.query.search_txt = dijit.byId('search_fruit').get('value');
grid.selection.clear();
grid.store.close();
grid._refresh();
}
How to achieve the same using the tree ? Thanks !
The refreshing of a dijit.Tree becomes a little more complicated, since there is a model involved (which in grid afaik is inbuilt, the grid component implements query functionality)
Performing search via store
But how to search, thats incredibly easy whilst using the ItemFileReadStore. Syntax is as such:
myTree.model.store.fetch({
query: {
name: 'Oranges'
},
onComplete: function(items) {
dojo.forEach(items, function(item) {
console.log(myTree.model.store.getValue(item, "ID"));
});
}
});
Displaying search results only
As shown above, the store will fetch, the full payload is put into its _allItemsArray and the store queryengine then filters out what its told by query argument to the fetch method. At any time, we could call fetch on store, even without sending an XHR for json contents - fetch with query argument can be considered as a simple filter.
It becomes slightly more interesting to let the Model know about this query.. If you do so, it will only create treeNodes to fill the tree, based on the returned results from store.fetch({query:model.query});
So, instead of sending store.fetch with a callback, lets _try to set model query and update the tree.
// seing as we are working with a multi-parent tree model (ForestTree), the query Must match a toplevel item or else nothing is shown
myTree.model.query = { name:'Fruits' };
// below method must be implemented to do so runtime
// and note, that the DnD might become invalid
myTree.update();
Refreshing tree with new xhr-request from store
You need to do exactly as you do with regards to the store. Close it but then rebuild the model. Model contains all the TreeNodes (beneath its root-node) and the Tree itself maps an itemarray which needs to be cleared to avoid memory leakage.
So, performing following steps will rebuild the tree - however this sample does not take in account, if you have DnD activated, the dndSource/dndContainer will still reference the old DOM and thereby 'keep-alive' the previous DOMNode hierachy (hidden ofc).
By telling the model that its rootNode is UNCHECKED, the children of it will be checked for changes. This in turn will produce the subhierachy once the tree has done its _load()
Close the store (So that the store will do a new fetch()).
this.model.store.clearOnClose = true;
this.model.store.close();
Completely delete every node from the dijit.Tree
delete this._itemNodesMap;
this._itemNodesMap = {};
this.rootNode.state = "UNCHECKED";
delete this.model.root.children;
this.model.root.children = null;
Destroy the widget
this.rootNode.destroyRecursive();
Recreate the model, (with the model again)
this.model.constructor(this.model)
Rebuild the tree
this.postMixInProperties();
this._load();
Creds; All together as such, scoped onto the dijit.Tree:
new dijit.Tree({
// arguments
...
// And additional functionality
update : function() {
this.model.store.clearOnClose = true;
this.model.store.close();
delete this._itemNodesMap;
this._itemNodesMap = {};
this.rootNode.state = "UNCHECKED";
delete this.model.root.children;
this.model.root.children = null;
this.rootNode.destroyRecursive();
this.model.constructor(this.model)
this.postMixInProperties();
this._load();
}
});
I'm trying to create a show function which needs to access to two documents: The document in 'doc' reference and another document called 'users'
My function looks like:
function(doc,req){
var friends = doc.friends;
var listFriends = [];
for(int i = 0; i<friends.length; i++){
var phone = friends[i].phone;
if(users[phone] != "" ){
listFriends.push(users[phone]);
}
}
return JSON.stringify(listFriends);
}
I'm not an expert nor javascript neither couchdb. My question is, Is it possible to access to the second document (users) in a similar way like in the code? So far it returns a compilation error.
Thanks
You can only access one document in a CouchDB show function. You could look at using a list function, which works on view results instead of documents.
Create a view where the two documents collate together (appear side-by-side in the view order) and you achieve an effect pretty close to what you wanted to achieve with the show function.