Asked  7 Months ago    Answers:  5   Viewed   60 times

I have scripts calling other script files but I need to get the filepath of the file that is currently running within the process.

For example, let's say I have three files. Using execfile:

  • script_1.py calls script_2.py.
  • In turn, script_2.py calls script_3.py.

How can I get the file name and path of script_3.py, from code within script_3.py, without having to pass that information as arguments from script_2.py?

(Executing os.getcwd() returns the original starting script's filepath not the current file's.)

 Answers

78

p1.py:

execfile("p2.py")

p2.py:

import inspect, os
print (inspect.getfile(inspect.currentframe()) # script filename (usually with path)
print (os.path.dirname(os.path.abspath(inspect.getfile(inspect.currentframe())))) # script directory
Tuesday, June 1, 2021
 
Sufi
answered 7 Months ago
40

Make sure you understand the three return values of os.walk:

for root, subdirs, files in os.walk(rootdir):

has the following meaning:

  • root: Current path which is "walked through"
  • subdirs: Files in root of type directory
  • files: Files in root (not in subdirs) of type other than directory

And please use os.path.join instead of concatenating with a slash! Your problem is filePath = rootdir + '/' + file - you must concatenate the currently "walked" folder instead of the topmost folder. So that must be filePath = os.path.join(root, file). BTW "file" is a builtin, so you don't normally use it as variable name.

Another problem are your loops, which should be like this, for example:

import os
import sys

walk_dir = sys.argv[1]

print('walk_dir = ' + walk_dir)

# If your current working directory may change during script execution, it's recommended to
# immediately convert program arguments to an absolute path. Then the variable root below will
# be an absolute path as well. Example:
# walk_dir = os.path.abspath(walk_dir)
print('walk_dir (absolute) = ' + os.path.abspath(walk_dir))

for root, subdirs, files in os.walk(walk_dir):
    print('--nroot = ' + root)
    list_file_path = os.path.join(root, 'my-directory-list.txt')
    print('list_file_path = ' + list_file_path)

    with open(list_file_path, 'wb') as list_file:
        for subdir in subdirs:
            print('t- subdirectory ' + subdir)

        for filename in files:
            file_path = os.path.join(root, filename)

            print('t- file %s (full path: %s)' % (filename, file_path))

            with open(file_path, 'rb') as f:
                f_content = f.read()
                list_file.write(('The file %s contains:n' % filename).encode('utf-8'))
                list_file.write(f_content)
                list_file.write(b'n')

If you didn't know, the with statement for files is a shorthand:

with open('filename', 'rb') as f:
    dosomething()

# is effectively the same as

f = open('filename', 'rb')
try:
    dosomething()
finally:
    f.close()
Saturday, June 5, 2021
 
TheLovelySausage
answered 6 Months ago
17

Try below code:

Resources resources = context.getResources();
int resourceId = resources.getIdentifier("navigation_bar_height", "dimen", "android");
if (resourceId > 0) {
    return resources.getDimensionPixelSize(resourceId);
}
return 0;
Sunday, June 6, 2021
 
Null
answered 6 Months ago
82

There are a few ways:

  • $0 is the currently executing script as provided by POSIX, relative to the current working directory if the script is at or below the CWD
  • Additionally, cwd(), getcwd() and abs_path() are provided by the Cwd module and tell you where the script is being run from
  • The module FindBin provides the $Bin & $RealBin variables that usually are the path to the executing script; this module also provides $Script & $RealScript that are the name of the script
  • __FILE__ is the actual file that the Perl interpreter deals with during compilation, including its full path.

I've seen the first three ($0, the Cwd module and the FindBin module) fail under mod_perl spectacularly, producing worthless output such as '.' or an empty string. In such environments, I use __FILE__ and get the path from that using the File::Basename module:

use File::Basename;
my $dirname = dirname(__FILE__);
Saturday, June 12, 2021
 
aaronhuisinga
answered 6 Months ago
39

Here's the simplest way to do it:

>>> import win32api
>>> win32api.GetLongPathName(win32api.GetShortPathName('texas.txt')))
'TEXAS.txt'
Wednesday, August 4, 2021
 
Tom Lilletveit
answered 4 Months ago
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