[Digikam-users] How to use versioning?

[Marcel Wiesweg]

> I'd like to use versioning in digiKam 2.6.0 but I have some problems.
> I've googled some documentation but if I repeat those steps here the
> result is different.
> 
> First I "Enabled Non-Destructive Editing and Versioning". I also
> played with "Always show original images" and "Always show
> intermediate snapshots" but they made no difference.
> 
> Then I edited a image in the image editor. I only made a single change
> (converted to black and white) and save as new version. Now the image
> editor shows me the new image in the "Image History" and the used
> filter in "Used Filters" in the "Versioning" sidebar. Back in digiKam
> album view I can see both images but no matter which one I select I
> can't see the complete image history in the versioning sidebar, it
> only shows the current selected image. "Used Filters" is empty, too.
> After that I selected both versions of the image and ran "Group
> selected here". Still no change in the versioning sidebar.
> I wonder if this is the expected behavior or what I'm doing wrong.

It must be right in the sidebar, toggling "Always show..." depends on 
identical information, and grouping is unrelated.
So I tried to reproduce the original problem (take one image, b&w, save as new 
version), but all information is available for me in the versioning sidebar.
If you can reliably reproduce it with 2.6.0, please take a (small) image, 
reproduce the bug, create a report on b.k.o, and attach original and result 
files for testing.


> Another thing I'm wondering is the shown image of grouped images. It
> seems to depend on where one runs "Group selected here". Shouldn't
> this depend on "Always show original images"? Or what is that setting
> for?

Dont confuse grouping with versioning. It's not directly related.
Yes, if you select A+B and choose "Group select HERE" on image A, then A will 
be group leader; if you run it on image B, then B will be group leader. As 
sometimes you may want to group on A, and sometimes on B.