How do I retrieve the name and path of the project selected? (Package Explorer)
example: c:\project\test\projectName
someone has some code that explains how to complete I get the project name or full path of a particular project in my workspace?
Eclipse defines an extension point
"org.eclipse.ui.navigator.linkHelper"
If you contribute a class to these EP you have to implement ILinkHelper
The ILInkHelper interface notifies you, when something was selected in the explorer
public void activateEditor(IWorkbenchPage aPage, IStructuredSelection aSelection)
You can check the type of the selection
if (aSelection.getFirstElement() instanceof IFile) {
// Do something
}
Old memories but maybe useful for you. I guess package explorer provides its selection, so you can get the current selection in your code by calling:
ISelectionService service = getSite().getWorkbenchWindow().getSelectionService()
than you can get the package explorer view by its id (plugin.xml for more details):
IStructuredSelection selection = (IStructuredSelection) service.getSelection("org.eclipse.jdt.ui.PackageExplorer");
Please note AFAIK you can always safely cast ISelection to IStructuredSelection. Then call structured.getFirstElement() and I think the first element will be an IFile object. I hope my "pseudo code" whould be enough for you. And IFile has lots of usefule methods for your convenience
Related
I am developing a eclipse plugin to ease the development using a proprietary version control system.
Right now there is only a command prompt version of the system available for this VCS and its run in terminal. So from my eclipse plugin I want to provide a simple menu options to do the things like check-out and check-in and internally call these commands.
But to run these commands I need to pass the argument 'path' of the selected .java file in the editor/project explorer. How can I get the path of the source file to the plugin?
Get the current workbench page:
IWorkbenchPage page = PlatformUI.getWorkbench().getActiveWorkbenchWindow().getActivePage();
the workbench page implements ISelectionService so you can get the current selection:
ISelection selection = page.getSelection();
this will generally by an IStructuredSelection (but you need to check)
IStructuredSelection sel = (IStructuredSelection)selection;
See if this adapts to an IFile:
Object selObject = sel.getFirstElement(); // or iterate through all the selection elements
IFile file = Platform.getAdapterManager().getAdapter(selObject, IFile.class);
if you get a file the full path is:
String location = file.getLocation().toOSString();
If the current part is an editor then the selection you receive might be a text string. So you need to deal with editors separately:
IWorkbenchPart activePart = page.getActivePart();
if (activePart instanceof IEditorPart)
{
IEditorInput input = ((IEditorPart)activePart).getEditorInput();
if (input instanceof IFileEditorInput)
{
IFile file = ((IFileEditorInput)input).getFile();
...
}
}
I writing my own text editor plugin for eclipse. I am now working on my own formatter. Actually, following that link http://wiki.eclipse.org/FAQ_How_do_I_support_formatting_in_my_editor%3F.
I have written my Strategy, I have overridden getContentFormatter in my SourceViewerConfiguration..
As I run my plugin and press Ctrl+Shift+F - and nothing happens.
I think that I'm missing a step here. Should I create a handler or something?
Thanks
Might it be you skipped the last part of the linked page?
Finally, you will need to create an action that invokes the formatter. No generic formatting action is defined by the text infrastructure, but it is quite easy to create one of your own. The action’s run method can simply call the following on the source viewer to invoke the formatter:
sourceViewer.doOperation(ISourceViewer.FORMAT);
What helped me. I have created a handler with the following executors body:
//get the editorPart
if (editorPart != null) {
ITextOperationTarget target = (ITextOperationTarget) editorPart
.getAdapter(ITextOperationTarget.class);
if (target instanceof ISourceViewer) {
ISourceViewer textViewer = (ISourceViewer) target;
((ITextOperationTarget) textViewer)
.doOperation(ISourceViewer.FORMAT);
}
}
Then just create menu items and bind them to the handler.
I want to get the path of current selected file in Eclipse workspace but my project is a simple view plug-in project.
I just want to display the name/path of the file selected as soon as user opens the view.
You get the current selection as mentioned by #Danail Nachev. See http://www.eclipse.org/articles/Article-WorkbenchSelections/article.html for information on working with the selection service.
Once you have the selection, the most common pattern is:
if (selection instanceof IStructuredSelection) {
IStructuredSelection ssel = (IStructuredSelection) selection;
Object obj = ssel.getFirstElement();
IFile file = (IFile) Platform.getAdapterManager().getAdapter(obj,
IFile.class);
if (file == null) {
if (obj instanceof IAdaptable) {
file = (IFile) ((IAdaptable) obj).getAdapter(IFile.class);
}
}
if (file != null) {
// do something
}
}
Edit:
Usually you get an InputStream from the IFile and process it that way. Using some FileSystemProviders or EFS implementations, there might not be a local path to the file.
PW
You can retrieve the current workbench selection using two methods with the following code:
through the Workbench SelectionService
getViewSite().getSelectionProvider().getSelection()
getViewSite().getWorkbenchWindow().getSelectionService()
You can find more information in this article.
Using the global workbench selection is the better approach, because it enables your view to get selection from everywhere, which is something the user might expect (I at least do). Also, almost all of the views in Eclipse (and I don't know exceptions to this rule) are using this approach.
If you absolutely must to tie your view to another view, then you can get all IWorkbenchPage iterate over them and search for the view by its ID and when you find the view, you call get its SelectionProvider to get the selection.
Only by reading this explanation, makes my hair go straight up. Considering that there might be multiple instances of the same view, you can end up with a selection from a random view. I'm not sure whether the user will make sense of how things work, unless you have some clear rules, which exactly view you need. It is up to you.
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Given the name ch/mollusca/sample/snippet.xml, is there an easy way to get a hold of this file in JDT code, when it is located in the projects classpath either as a source file or inside a JAR?
The file may also be inside another project, that is referenced by the one, where I'm trying to get a hold of the actual file behind the name.
This is specific to Java projects, so it's possible to get a hold of the IJavaProject if that helps.
This is probably not the answer you want, but it's the best I can do (and since your question already has 2 days, I might has well throw you this):
Eclipse Corner Articles: Abstract Syntax Tree
Eclipse JDT - Abstract Syntax Tree (AST) and the Java Model - Tutorial
These two great articles will point you to the right direction.
How about this:
IEditorPart activeEditor = PlatformUI.getWorkbench().getActiveWorkbenchWindow().getActivePage().getActiveEditor();
if ( activeEditor != null ) {
IResource activeEditorResource = (IResource)activeEditor.getEditorInput().getAdapter( IResource.class );
if ( activeEditorResource != null && activeEditorResource.getFullPath() != null ) {
String activeFileName = activeEditorResource.getFullPath().toOSString();
//...do something with the active file
}
}
I am writing Eclipse plugins for Java, and have the following problem:
Given an IEditorPart, I need to check if it is a java editor.
I could do (IEditor instanceof JavaEditor),
but JavaEditor is an org.eclipse.jdt.internal.ui.javaeditor.JavaEditor,
which falls under the JDT's "internal" classes.
Is there a smarter and safer way to do this? I'm not sure why there is no non-internal interface for this.
You should test the id of the IEditorPart:
private boolean isJavaEditor(IWorkbenchPartReference ref) {
if (ref == null) {
return false; }
String JavaDoc id= ref.getId();
return JavaUI.ID_CF_EDITOR.equals(id) || JavaUI.ID_CU_EDITOR.equals(id);
}
Testing the instance was only needed in eclipse3.1.
alt text http://blogs.zdnet.com/images/Burnette_DSCN0599.JPG
JavaUI is the main access point to the Java user interface components. It allows you to programmatically open editors on Java elements, open a Java or Java Browsing perspective, and open package and type prompter dialogs.
JavaUI is the central access point for the Java UI plug-in (id "org.eclipse.jdt.ui")
You can see that kind of utility function ("isJavaEditor()") used for instance in ASTProvider.
The mechanism of identification here is indeed simple String comparison.
Anyway, you are wise to avoid cast comparison with internal class: it has been listed as one of the 10 common errors in plugins development ;) .
One strategy might be to use JavaUI.getEditorInputJavaElement(IEditorPart):
// given IEditorPart editor
IJavaElement elt = JavaUI.getEditorInputJavaElement(editor.getEditorInput());
if (elt != null) {
// editor is a Java editor
}
The method returns null if the editor input is not in fact a Java element.