Passing a string variable to glob - python-3.x

I'm trying to pass a directory to glob which will be read to a variable from a config file.
If I just do this it works fine:
path = '//Server/Company/Official Documents/**/*.pdf'
Files = glob.glob(path,recursive=True)
But if I try to do this I get an empty list:
path = Config[1][1]
Files = glob.glob('{path}**/*.pdf'.format(path=path),recursive=True)
For information,
print(Config [1][1])
gives this
'//Server/Company/Official Documents/'

You are missinng / in '{path}**/*.pdf'.format(path=path)
Try changinng it to:
'{path}/**/*.pdf'.format(path=path1)
like:
glob.glob('{path}/**/*.pdf'.format(path=path),recursive=True)

Related

Striping a string from newline markers

I store configuration data (paths to specific files) inside a file named app.cfg that looks like this :
path/to/config.json
path/to/default/folder
and I query those item with the following Python code:
with open("app.cfg","r",newline='') as config:
data = config.readlines()
PathToConfig = data[0]
DefaultPath = data[1]
config.close()
But when I use PathToConfig in my script, the path stored in this variable cannot be used because there is \n at the end of the string.
I tried to fix this issue by using this PathToConfig = data[0].rstrip() but there still is \n at the end of the string.
How can I strip this string from the newline marker ?
You should be able to solve it with .rstrip to strip "\n":
create app.cfg:
with open("app.cfg","w",newline='') as config:
config.writelines("""path/to/config.json
path/to/default/folder""")
app.cfg looks like this:
read contents from file:
with open("app.cfg","r",newline='') as config:
data = config.readlines()
PathToConfig = data[0].rstrip("\n")
DefaultPath = data[1]
output:

Python Glob - Get Full Filenames, but no directory-only names

This code works, but it's returning directory names and filenames. I haven't found a parameter that tells it to return only files or only directories.
Can glob.glob do this, or do I have to call os.something to test if I have a directory or file. In my case, my files all end with .csv, but I would like to know for more general knowledge as well.
In the loop, I'm reading each file, so currently bombing when it tries to open a directory name as a filename.
files = sorted(glob.glob(input_watch_directory + "/**", recursive=True))
for loop_full_filename in files:
print(loop_full_filename)
Results:
c:\Demo\WatchDir\
c:\Demo\WatchDir\2202
c:\Demo\WatchDir\2202\07
c:\Demo\WatchDir\2202\07\01
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_51.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_52.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_53.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_54.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_55.csv
c:\Demo\WatchDir\2202\07\05
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_00.csv
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_01.csv
Results needed:
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_51.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_52.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_53.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_54.csv
c:\Demo\WatchDir\2202\07\01\polygonData_2022_07_01__15_55.csv
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_00.csv
c:\Demo\WatchDir\2202\07\05\polygonData_2022_07_05__12_01.csv
For this specific program, I can just check if the file name contains.csv, but I would like to know in general for future reference.
Line:
files = sorted(glob.glob(input_watch_directory + "/**", recursive=True))
replace with the line:
files = sorted(glob.glob(input_watch_directory + "/**/*.*", recursive=True))

Creating an empty folder on Dropbox with Python. Is there a simpler way?

Here's my sample code which works:
import os, io, dropbox
def createFolder(dropboxBaseFolder, newFolder):
# creating a temp dummy destination file path
dummyFileTo = dropboxBaseFolder + newFolder + '/' + 'temp.bin'
# creating a virtual in-memory binary file
f = io.BytesIO(b"\x00")
# uploading the dummy file in order to cause creation of the containing folder
dbx.files_upload(f.read(), dummyFileTo)
# now that the folder is created, delete the dummy file
dbx.files_delete_v2(dummyFileTo)
accessToken = '....'
dbx = dropbox.Dropbox(accessToken)
dropboxBaseDir = '/test_dropbox'
dropboxNewSubDir = '/new_empty_sub_dir'
createFolder(dropboxBaseDir, dropboxNewSubDir)
But is there a more efficient/simpler way to do the task ?
Yes, as Ronald mentioned in the comments, you can use the files_create_folder_v2 method to create a new folder.
That would look like this, modifying your code:
import dropbox
accessToken = '....'
dbx = dropbox.Dropbox(accessToken)
dropboxBaseDir = '/test_dropbox'
dropboxNewSubDir = '/new_empty_sub_dir'
res = dbx.files_create_folder_v2(dropboxBaseDir + dropboxNewSubDir)
# access the information for the newly created folder in `res`

How to get the name of the directory from the name of the directory + the file

In an application, I can get the path to a file which resides in a directory as a string:
"/path/to/the/file.txt"
In order to write another another file into that same directory, I want to change the string "/path/to/the/file.txt" and remove the part "file.txt" to finally only get
"/path/to/the/"
as a string
I could use
string = "/path/to/the/file.txt"
string.split('/')
and then glue all the term (except the last one) together with a loop
Is there an easy way to do it?
You can use os.path.basename for getting last part of path and delete it with using replace.
import os
path = "/path/to/the/file.txt"
delete = os.path.basename(os.path.normpath(path))
print(delete) # will return file.txt
#Remove file.txt in path
path = path.replace(delete,'')
print(path)
OUTPUT :
file.txt
/path/to/the/
Let say you have an array include txt files . you can get all path like
new_path = ['file2.txt','file3.txt','file4.txt']
for get_new_path in new_path:
print(path + get_new_path)
OUTPUT :
/path/to/the/file2.txt
/path/to/the/file3.txt
/path/to/the/file4.txt
Here is what I finally used
iter = len(string.split('/'))-1
directory_path_str = ""
for i in range(0,iter):
directory_path_str = directory_path_str + srtr.split('/')[i] + "/"

How to get the default application mapped to a file extention in windows using Python

I would like to query Windows using a file extension as a parameter (e.g. ".jpg") and be returned the path of whatever app windows has configured as the default application for this file type.
Ideally the solution would look something like this:
from stackoverflow import get_default_windows_app
default_app = get_default_windows_app(".jpg")
print(default_app)
"c:\path\to\default\application\application.exe"
I have been investigating the winreg builtin library which holds the registry infomation for windows but I'm having trouble understanding its structure and the documentation is quite complex.
I'm running Windows 10 and Python 3.6.
Does anyone have any ideas to help?
The registry isn't a simple well-structured database. The Windows
shell executor has some pretty complex logic to it. But for the simple cases, this should do the trick:
import shlex
import winreg
def get_default_windows_app(suffix):
class_root = winreg.QueryValue(winreg.HKEY_CLASSES_ROOT, suffix)
with winreg.OpenKey(winreg.HKEY_CLASSES_ROOT, r'{}\shell\open\command'.format(class_root)) as key:
command = winreg.QueryValueEx(key, '')[0]
return shlex.split(command)[0]
>>> get_default_windows_app('.pptx')
'C:\\Program Files\\Microsoft Office 15\\Root\\Office15\\POWERPNT.EXE'
Though some error handling should definitely be added too.
Added some improvements to the nice code by Hetzroni, in order to handle more cases:
import os
import shlex
import winreg
def get_default_windows_app(ext):
try: # UserChoice\ProgId lookup initial
with winreg.OpenKey(winreg.HKEY_CURRENT_USER, r'SOFTWARE\Microsoft\Windows\CurrentVersion\Explorer\FileExts\{}\UserChoice'.format(ext)) as key:
progid = winreg.QueryValueEx(key, 'ProgId')[0]
with winreg.OpenKey(winreg.HKEY_CURRENT_USER, r'SOFTWARE\Classes\{}\shell\open\command'.format(progid)) as key:
path = winreg.QueryValueEx(key, '')[0]
except: # UserChoice\ProgId not found
try:
class_root = winreg.QueryValue(winreg.HKEY_CLASSES_ROOT, ext)
if not class_root: # No reference from ext
class_root = ext # Try direct lookup from ext
with winreg.OpenKey(winreg.HKEY_CLASSES_ROOT, r'{}\shell\open\command'.format(class_root)) as key:
path = winreg.QueryValueEx(key, '')[0]
except: # Ext not found
path = None
# Path clean up, if any
if path: # Path found
path = os.path.expandvars(path) # Expand env vars, e.g. %SystemRoot% for ext .txt
path = shlex.split(path, posix=False)[0] # posix False for Windows operation
path = path.strip('"') # Strip quotes
# Return
return path

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