how to get output file path with shutil.copy

Question

import tempfile
import shutil

temp_ =  tempfile.mkdtemp()
class ListView_(Screen):
    def Image_(self, path):
        global image_file_path
        file_path = shutil.copy2(path[0], temp_)

in python 3 file_path output is a "path"

in python 2 file_path output is "None" so how to get the path of the new file in temp directory

Answer 1

shutil comes with source, so you can look into python 3 version and adapt it.

If you compare shutil.copy2 methods for Python 2.7 & 3.4, you'll notice a new return dst in the 3.4 version. That's a new feature added, not present in python 2.

the interesting lines for you in shutil.copy2 method are:

if os.path.isdir(dst):
    dst = os.path.join(dst, os.path.basename(src))

It means that if temp_ is a directory, then target is the directory / basename of the source, else leave it as is, so after this code, dst is always a target file name, and open(dst,"wb') will work.

so to make your code compatible with python 2 & 3 (which is a good thing when it's possible), you can emulate that by using a ternary expression to compute the actual filepath target, then use it directly in shutil (why passing a directory again ?):

file_path = os.path.join(_temp, os.path.basename(src)) if os.path.isdir(temp_) else temp_
shutil.copy2(path[0], file_path)  # ignore return code