使用 Python 打开 gimp 文件夹中的所有图像

Question

I get something similar using:

gimp = r'C:\Program Files\GIMP 2\bin\gimp-2.10.exe'

os.chdir(folder)

filesnames= os.listdir()

for f in filesnames:
    actfile = folder+'\\'+f
    p = subprocess.Popen([gimpcamin, actfile])
    p.wait()

But this script can open only one image and the next image opens when I close Gimp.

Note: I am using VSCode

Answer 1

I believe that the issue is that you have p.wait() immediately after p = subprocess.Popen([gimpcamin, actfile]) , which makes it wait for that command (ie, Gimp) to return before proceeding.

You should try:

gimp = r'C:\Program Files\GIMP 2\bin\gimp-2.10.exe'

os.chdir(folder)

filesnames= os.listdir()

processes = []
for f in filesnames:
    actfile = folder+'\\'+f
    processes.append(subprocess.Popen([gimpcamin, actfile]))

# wait for all processes to finish
for p in processes:
    p.wait()

I'm not sure that this what you really want. I suspect you rather have just another Window of Gimp open for each image, rather than opening a whole different instance. You should double-check Gimp's interface to see if that's possible.

Answer 2

If this is all you do, you can just use gimp /path/dir/*.png in your shell, like most Unix utilities, the gimp command accepts several files.

If you want to do it in Python:

• Since you do a chdir(), Gimp's current directory when you call it is the directory to which you moved, so you can use the path-less names obtained with listdir()
• You can pass all the files to Gimp in one call so something like:

gimpCmd=[r'C:\Program Files\GIMP 2\bin\gimp-2.10.exe','-n'] # -n to get a new Gimp instance, if necessary
subprocess.Popen(gimpCmd+os.listdir())

Of course this is possibly limited by the acceptable command line length, but in practice you'll be out of RAM (and display real estate) before you hit that limit.