[PyQt] Layout and widget interaction

[David Boddie]

On Fri, 14 May 2010 22:37:52 +0530, Anshul Jain wrote:

> I have made a very small trial application in which there are 2 widgets
> inside a horizontal layout.(Please see attached image). One contains images
> and the other contains a text editor. The code snippet that adds these
> widgets to this layout is:
>
> mainWidget = QtGui.QWidget()
>     horizontalLayout = QtGui.QHBoxLayout()
>     horizontalLayout.addWidget(ImageWidget())
>     horizontalLayout.addWidget(editorWidget())
>
> My aim is that when i click on an image  in the ImageWidget() it must reads
> a file corresponding to that image and then display the text in
> editorWidget(). I have been able to write code to find which image has been
> l have clicked but then I don't know to transfer that filename to the
> editorWidget() in the horizontalLayout. How should I do it. Please help me
> out. I'm really stuck at it.  :(

I suggest making the ImageWidget() emit a signal that you receive in a slot
in the editorWidget(). You will need to connect the signal somewhere, and
it's probably best to do that before you lay out those widgets.

  imageWidget = ImageWidget()
  editor = editorWidget()
  imageWidget.clickedImage.connect(editor.openFile)

  mainWidget = QtGui.QWidget()
  horizontalLayout = QtGui.QHBoxLayout()
  horizontalLayout.addWidget(imageWidget)
  horizontalLayout.addWidget(editor)

In the image widget you will need to define a custom signal called
clickedImage that takes a string as an argument.

  clickedImage = QtCore.pyqtSignal(QtCore.QString)

You need to emit this when handling the click.

In the editor widget, you simply need to define a method called openFile,
like this:

  def openFile(self, path):
    # Open the file with the given path.

Hope this helps,

David