How to delete a layer in script-fu

Graechan · **Posted:** Tue May 31, 2011 9:22 pm **(#1)**

I have been left with these layers.

And I'd like to delete the "Brouillard uni" layer before displaying the image with (gimp-display-new img)
Could somebody help me with the correct procedure as Wilbur does not like anything I have tried.

GnuTux · **Posted:** Tue May 31, 2011 9:27 pm **(#2)**

Have you tried gimp-image-remove-layer?

Code: Select all

(gimp-image-remove-layer img layer_name)

PhotoComix · **Joined:** Apr 12, 2010 **Posts:** 5870

i am looking the PB from gimp 2.7.3.
it says that "gimp-layer-delete" is deprecated and to use instead " gimp-item-delete" that has as argoment the id of the "item"

EDIT
Ninja posted, but Gnu-Tux should know better and anyway better 2 options then none :mrgreen:

Graechan · **Posted:** Tue May 31, 2011 10:28 pm **(#4)**

I tried (gimp-image-remove-layer img "Brouillard uni") but wilber said invalid type to argument 2

Oregonian · **Posted:** Tue May 31, 2011 10:30 pm **(#5)**

The standard method of deleting a layer doesn't work?

GnuTux · **Posted:** Tue May 31, 2011 10:43 pm **(#6)**

Graechan wrote:

I tried (gimp-image-remove-layer img "Brouillard uni") but wilber said invalid type to argument 2

You need to use the pointer variable name (let* statement) you defined in the script , rather than the textual name in your layers dialog. You may have to save it so you can point back to it later and delete it.

PhotoComix · **Joined:** Apr 12, 2010 **Posts:** 5870

@ O
Graechan is asking how delete a layer in a script-fu

@Graechan
because bruillard uni is the layer name but not the layer ID
I didn't like much script fu also for this type of obstacles, ...you cannot just order to delete the "top" or, the "bottom" layer or the active layer...first you have to recover the ID assigned by gimp to the layer...that has no relation with the layer name

But..sorry i forgot how to, i only remember that was explained in one of the script fu tutorial by FencePost on http://www.gimpdome.com

@Gnu Tux
Ninja posted again ! :lol

Still i believe the layer name passed as paramether instead then the layer ID remain a problem

Graechan · **Posted:** Wed Jun 01, 2011 12:38 am **(#8)**

Fencepost wrote of using the "car" command to assign a layer ID so I tried this.

(set! activelayer (car (gimp-image-set-active-layer img "Brouillard uni")))

But i still get the same error on this command

Oregonian · **Posted:** Wed Jun 01, 2011 12:49 am **(#9)**

PhotoComix wrote:

@ O
Graechan is asking how delete a layer in a script-fu

Oh, my, how embarrassing. Forget I did that. Sheesh!

mahvin · **Joined:** Oct 06, 2010 **Posts:** 4050

Layer IDs are like definition variables (no quotes). Look at the define variables, and then look at the let* statements. You'll see layers being referred to things like:

text-layer

In this case, GnuTux pointed out that there is a let* statement that defines this layer and he suggested you use the procedure he recommends with that layer variable not the layer name. Make sense?

Graechan · **Posted:** Wed Jun 01, 2011 1:22 am **(#11)**

TA-DA

I got it

I had previously assigned "brouillard uni" as "activelayer" in the Let block all I needed was (gimp-image-remove-layer img activelayer). which did not occur to me because Frame 9 was the active layer.

Thanks for prodding me Mahvin

paynekj · **Posted:** Wed Jun 01, 2011 1:30 am **(#12)**

Agh I got Ninja'd.....

But here's what I was writing any way ...

The delete procedure will be expecting an id number not a string, so you need to find the id of the "Brouillard uni" layer. Here's some code I wrote as part of someone else's problem that finds a layer id based on it's name:

Code: Select all

; routine to find the Id of a layer given the layer name
(define (kp24_find_layer_by_name image layerName)
  (let* (
           (layerList (vector->list (cadr (gimp-image-get-layers image))))    ; get the list of layer ids
           (wantedLayerId -1)
           (layerText "")
        )
        
        (for-each
          (lambda (layerId)
            (set! layerText (car (gimp-drawable-get-name layerId))) ; get a layer name
            (if (string=? layerText layerName)   ; is it the one we want?
              (set! wantedLayerId layerId)       ; it is, so save the layer id
            )            
          )
          layerList
        )
        (if (= -1 wantedLayerId)      ; didn't find it so complain
          (error (string-append "Could not find a layer with name:- " layerName))
        )
        (list wantedLayerId)
  )
)

to use it you pass it the image id and the string that is name of your layer. It returns a list containing the id of the layer:

Code: Select all

   (set! wanted_layer_id (car (kp24_find_layer_by_name imageId "Brouillard uni")))
   (gimp-drawable-delete wanted_layer_id)

Kevin

ofnuts · **Joined:** Oct 25, 2010 **Posts:** 4812

I don't use script-fu, but another possibility is to iterate the layers and delete those with a name that match "Frame*" (or whatever regexp will work in TinyScheme). Personally, I make a habit of calling my temporary work items (layers, gradients, brushes...) with a "#" so that they are easier to spot.

In python-fu you have an <Image> object, that has an array of <Layer> (unexpectedly called "layers"), ordered as in the layer stack, so the top layer is image.layers[0]. Welcome to the 21st century

PS: sorta ninja'd by paynejk myself

Graechan · **Posted:** Wed Jun 01, 2011 3:25 am **(#14)**

thank you Kevin I noted all that in case I need in future. And Offnuts its evident I don't do script-fu either :hehe