How to open Windows Explorer on a selected resource in Eclipse

Question

I was looking for a small plug-in for Eclipse that will open Windows Explorer on the currently selected resource from the package explorer tree.

I know that Aptana Studio provides this functionality through the context menu on the resource, but it has many other things that are not interesting to me.

Are there other solutions?

Answer 1

I use this plugin seems ok

New link for Eclipse update https://fabioz.github.com/startexplorer/update/

Old reference link

• http://sourceforge.net/projects/startexplorer/ *

• http://basti1302.github.com/startexplorer/update/

Answer 2

In Eclipse Luna and then select a resource and then:

Alt + shift + W > System Explorer

or

Right-click> Show in> System Explorer

Here you can configure the exact command that must be executed to open System Explorer:

Window> Preferences> General> Workspace> System Explorer Launch Command

Answer 3

In fact, you can do this through the built-in external tool manager. Here are the instructions: http://www.eclipsezone.com/eclipse/forums/t77655.html I'm trying to get it to work with Nautilus. However, it works under Windows when I tried it.

Answer 4

StartExplorer does not work under my Ubuntu, but ExploreFS works. You can find it here:

http://junginger.biz/eclipse/

It supports Windows, Mac OS X and Linux.

Answer 5

Download the jar OpenExplorer file. I am using OpenExplorer_1.5.0.v201108051513.jar, downloaded from https://github.com/samsonw/OpenExplorer/archives/master .

Copy this to your eclipse / plugins folder and restart Eclipse. It is very convenient. I would recommend eclipse users having this.

Answer 6

Eclipse Explorer is an eclipse plugin that helps you quickly open a folder or select a resource in Explorer. It supports key help, can open the shared resource and the entire location of java elements, even .jar in the library.

Additional function:

• Support for all java element explorer
• Support Key Assistant (default Ctrl + ` )
• Support for Windows and Linux platforms.
• Support for automatic file selection (Windows only)

Answer 7

Create a new plug-in project using the Eclipse PDE. Connect your Activator class to the Common Navigator API to get a choice for IResource . For each IResource selected IResource use FileLocator to get the file URI with which you can create a java.io.File object. Then it can be opened in the explorer's own files explorer using the integration with Java 6 Desktop :

  if (Desktop.isDesktopSupported()) { Desktop desktop = Desktop.getDesktop(); desktop.open(new File("C:/")); } 

Answer 8

 open explorer in eclipse - in eclipse -> external tools configurations - in program tree -> new name: OpenExplore localtion: C:\Windows\explorer.exe Arguments: /select,${selected_resource_loc}\ 

Answer 9

The default Linux command (dbus-send ...) fails on CentOS 6 and CentOS 7. Changing it to nautilus "${selected_resource_parent_loc}" makes it work. I got this information from this documentation page , which I received from this error report .

I am creating an RCP application and I do not want my users to manually change this setting. Using a spyware spy, I found an appropriate repository of preferences and a key. So this non-API call will programmatically set preference:

  IDEWorkbenchPlugin.getDefault().getPreferenceStore().setValue(IDEInternalPreferences.WORKBENCH_SYSTEM_EXPLORER, "nautilus \"${selected_resource_parent_loc}\""); 

In newer versions of nautilus, you can specify ${selected_resource_loc} instead, in which case it opens the parent folder with the selected resource. I watched this with nautilus v 3.14, but in version 2.28 an error is thrown: the resource is not a folder.

Answer 10

An easy way to open a directory in Windows Explorer is to select a file in Project Explorer, press Alt + Shift + W and press X.

Answer 11

I use the EasyShell plugin for Eclipse, it has this functionality and much more.

Look at this:

https://anb0s.imtqy.com/EasyShell/