I am currently developing an eclipse plugin which displays DOT-Graphs. For this purpose I make use of this plugin. However, I have no idea how to actually display the graph which I built. I want to display it in the middle of the eclipse window as an Editor.
To get this done I created a custom Editor class which needs some code in its createPartControl(Composite) code in order to make use of the DotGraphView which is provided by the plugin.
The question is, how can I display this DotGraphView?
The code of my Editor looks like this:
#Override
public void createPartControl(Composite container) {
DotImport importer = new DotImport(TEST_GRAPH);
Graph graph = importer.newGraphInstance();
DotGraphView dotGraphView = new DotGraphView();
dotGraphView.setGraph(graph);
// add dotGraphView as a child to container and display it
// What todo here?
}
To use the graph in your own custom view, check out the implementation of ZestFxUiView, the superclass of DotGraphView. You could probably subclass ZestFxUiView and call setGraph with your graph object.
Related
I'm trying to write a VSTO-Add-In with a System.Windows.Forms.WebBrowser-Control enabling something similar to the Office-JS-Add-In model.
The WebBrowser-control would show some HTML/JS-Page and be able to call C#-functions in the VSTO-Add-In from JavaScript via window.external and the ObjectForScripting-property of the WebBrowser-object.
That is in JS the call would be
window.external.DoFancyStuffToMyDocument(withTheseParams)
while there had to be some
class MyFunctionProxy() {
public void DoFancyStuffToMyDocument(string theParam) {
//code here
}
}
in the C#-Code an this would be attached to the WebBrowser
myWebBrowser.ObjectForScripting = new MyFunctionProxy();
So far so good. Now comes the catch. I want my HTML/JS-Code be able to also utilize the office.js code and functions like
Word.run(function (context) {
var thisDocument = context.document;
var range = thisDocument.getSelection();
range.insertText('"Hitch your wagon to a star."\n', Word.InsertLocation.replace);
//...
}
Does anyone see a way of getting this to work?
My initial guess was that the OfficeJS-taskpane-add-ins in Word on-prem use some some similar methode as above with a class derived from WebBrowser and the appropriate ObjectForScripting. This would then suggest that there must be a (hopefully accessible) class which is assigned to the ObjectForScripting-property handling the function calls from office.js. Then I could proxy this ObjectForScripting-class and add my own functions like 'DoFancyStuffToMyDocument()'.
I've written an Xtext-based plugin for some language. I'm now interested in creating a new independent view (as a separate plugin, though it requires my first plugin), which will interact with the currently-active DSL document - and specifically, interact with the model Xtext parsed (I think it's called the Ecore model?). How do I approach this?
I saw I can get an instance of XtextEditor if I do something like this when initializing my view:
getSite().getPage().addPartListener(new MyListener());
And then, in MyListener, override partActivated and partInputChanged to get an IWorkbenchPartReference, which is a reference to the XtextEditor. But what do I do from here? Is this even the right approach to this problem? Should I instead use some notification functionality from the Xtext side?
Found it out! First, you need an actual document:
IXtextDocument doc = editor.getDocument();
Then, if you want to access the model:
doc.modify(new IUnitOfWork.Void<XtextResource>() { // Can also use just IUnitOfWork
#Override public void process(XtextResource state) throws Exception {
...
}
});
And if you want to get live updates whenever it changes:
doc.addModelListener(new IXtextModelListener() {
#Override public void modelChanged(XtextResource resource) {
for (EObject model : resource.getContent()) {
...
}
}
});
I'm writing an Eclipse plugin, I want to create a wizard for my new project type. I created pages by classes extends org.eclipse.jface.wizard.WizardPage. My requirement is, based on some condition in one page, I need to go back to previous page without pressing back button on page(programmatically).
Is it possible?
Thanks a million in advance!
I don't think this is a good idea. The user will be confused by doing this. I would disable the finish and the next button and give an error, telling the user that he has to go back to the first page.
If you want to reuse some UI from the first page, define the UI as a new class and reuse it.
I implemented something similar using the WizardDialog:showPage() method:
MyWizard.java
public void createPageControls(Composite pageContainer) {
// TODO Auto-generated method stub
super.createPageControls(pageContainer);
wizardDialog = (WizardDialog) getContainer();
}
public void skipProcessPage() {
wizardDialog.showPage(workPage == arisDbPage ? focusPage : arisDbPage);
}
public void setWorkPage(IWizardPage workPage) {
this.workPage = workPage;
}
here the processPage does the lengthy db lookup!
HTH thomas
I am currently building a RCP application based on Eclipse.
In one of my plugins I am adding two views via code:
layout.addView("dev.asd.tableviewer.tree", IPageLayout.LEFT, 0.25f, IPageLayout.ID_EDITOR_AREA);
layout.addView("dev.asd.tableviewer.view", IPageLayout.RIGHT, 0.75f, IPageLayout.ID_EDITOR_AREA);
The first view contains a treeviewer, the second one a tableviewer. Now I want to update the tableviewer's content according to the selection of the treeviewer. My question is, how can I reference the tableviewer from within the treeviewer? Or is there an other way to solve this problem?
Use the SelectionService for this issue. No referencing of views required, see http://www.eclipse.org/articles/Article-WorkbenchSelections/article.html
In each view, define an ID. The ID is the same as what you defined in the ID field when you defined the view in the extension element details. Here's one of mine:
public static final String ID = "gov.bop.rabid.ui.views.PrefetchedInmatesView";
In your RCP plug-in, define the following method:
public static IViewPart getView(IWorkbenchWindow window, String viewId) {
IViewReference[] refs = window.getActivePage().getViewReferences();
for (IViewReference viewReference : refs) {
if (viewReference.getId().equals(viewId)) {
return viewReference.getView(true);
}
}
return null;
}
When you want to reference a view from a different view, use the following code:
PrefetchedInmatesView view = (PrefetchedInmatesView)
RabidPlugin.getView(window, PrefetchedInmatesView.ID);
Substitute the name of your view for PrefetchedInmatesView, and the name of your plug-in for RabidPlugin.
I have a login button in the header of the website. This header's html is programmed into Zend framework views/layouts/home.phtml.
I have a hidden form in this layout that is triggered by jQuery thickbox inline content display integration. Reason, I dont want to make a ajax call to just fetch a small login form.
I create the form using Zend_Form and the problem is that I have to do it in all the controllers after checking if the user is logged in or not. I want to place this form generation in one single place, say in bootstrap and then have a logic in bootstrap to say that if user is logged in dont generate the form.
I don't know if bootstrap is the right place to do so or should I do it in some other place.
So, where should I instantiate the form so that its available everywhere if user is not logged in.
Create your own base controller which extends Zend_Controller_Action then have your controllers extend off of your base controller. I don't know what "jQuery thickbox inline content display integration" is...but you have several sections you can put it in depending when you need your code to run. init(), preDispatch(), postDispatch() etc... Just make sure when you extend off your base controller that you do sthing like:
parent::init()
parent::preDispatch()
parent::postDispatch()
etc... within each section so that the base code runs as well...
Be careful about Pradeep Sharma's solution (the answer he wrote himself and accepted below).
All the code code below is for ZF 1.12, and not ZF 2.0
In the bootstrap, Zend_Layout's MVC instance might not have been created yet. You should use Zend_Layout::startMvc() instead :
$view = Zend_Layout::startMvc()->getView() ;
And tbh I prefer executing this code in the preDispatch() function. New users of ZF might be interested in this :
application/plugins/HeaderForm.php :
class Application_Plugin_HeaderForm extends Zend_Controller_Plugin_Abstract
{
public function preDispatch(Zend_Controller_Request_Abstract $request)
{
$view = Zend_Layout::startMvc()->getView() ;
$view->headerForm = new Application_Form_HeaderForm() ;
}
}
Calling new Application_Form_HeaderForm() will autoload by default into application/forms/ folder. You can also create the form directly into the plugin with new Zend_Form(), and addElement() etc. but it won't be reusable.
Of course, you need to register this plugin in your bootstrap!
application/Bootstrap.php :
class Bootstrap extends Zend_Application_Bootstrap_Bootstrap
{
protected function _initPlugin()
{
$front = Zend_Controller_Front::getInstance() ;
$front->registerPlugin(new Application_Plugin_HeaderForm()) ;
}
}
Calling new Application_Plugin_HeaderForm() will autoload by default into application/plugins/ folder
I did it in a different way, extendingZend_Controller_Plugin_Abstract to implement a plugin and register it with front controller.
public function routeStartup(Zend_Controller_Request_Abstract $request) { }
generated the form inside the above mentioned method and by setting the form in $view object.
$view can be retrived using :
$view = Zend_Layout :: getMvcInstance()->getView();