Cannot find file path startfile python

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August undefined, 2024

WebJul 3, 2016 · If you start the Python script from another directory, e.g. in a command prompt or the run dialog, then the working directory won't be the Python script directory, and startfile will fail. The VBS script also inherits this working directory. WebJan 16, 2013 · If this is a Python 3 & unicode problem, I suggest trying to fix the string first: path = "\\\\myshare\folder" path = bytes (path, "utf-8").decode ("unicode_escape") print …

windows - Python os.startfile Can

WebOct 29, 2024 · os.startfile require file directory + file name (if your script not in the same directory with the files) import random import os himg = 'C:\\Users\\Vl\\Desktop\\aaaa\\himg\\' files = os.listdir (himg) d = random.choice (files) rng1 = (random.randint (0, 10)) if (rng1 % 2) == 0: os.startfile (himg + d) Share Improve this … WebJul 19, 2024 · If os.startfile(lines_kml_flyingpath) raises FileNotFoundError, then it’s the KML file itself that can’t be found, as opposed to “open.exe” with the original subprocess … citi hardware price list pdf

python - os.startfile saying file doesn

WebFeb 26, 2024 · #!/usr/bin/python # -*- coding: utf-8 -*- import os import enum from os import path, startfile # Enum for size units class SIZE_UNIT (enum.Enum): BYTES = 1 KB = 2 MB = 3 GB = 4 def convert_unit (size_in_bytes, unit): """ Convert the size from bytes to other units like KB, MB or GB""" if unit == SIZE_UNIT.KB: return size_in_bytes/1024 elif unit … WebJun 2, 2024 · The system cannot find the file specified: 'PG2356E2-26.jpg' You are intending to operate on a path name (like "C:\PG2\356E2-26.jpg"), but are instead handing to rename () a string with all that mushed together. You didn't put any path separators in. You could do that manually, but better is to use os.path functions to form them. 1 2 3 4 5 … WebDec 19, 2024 · 1 I saw a question on here somewhat like mine, but the solution there did not work. My code is: for filename in os.scandir ('\\\\network_drive\\folder\\folder\\folder\\'): print (filename) The error is: FileNotFoundError: [WinError 67] The network name cannot be found: '\\\\network_drive\\folder\\folder\\folder\\' diashow 9 ultimate

python - How to open a .pdf file, knowing only part of it

WebYou cannot use an absolute path, unless your terminal is in that directory. Hence, you can do as in the following: import os def rename(directory): os.chdir(directory) # Changing to … WebMar 25, 2016 · Please make sure of the following: The parent directory of the folder (JSONFiles) is the same as the directory of the Python script. Even though the folder … diashow als bildschirmschonerWebOct 30, 2013 · There are lots of alternatives to fix this, starting with doubling the backslash: os.path.join is the safest and most portable choice. As long as you have "c:" hardcoded … citihardware pagadian

"WebMar 7, 2024 · file = r'c/:folder/file.txt' os.startfile (file) returns FileNotFoundError: [WinError 2] The system cannot find the file specified: 'c:/folder/file.txt' I have also tried to check if … " - Cannot find file path startfile python

Cannot find file path startfile python

python - Retrieving contents from a directory on a network drive ...

WebSep 9, 2008 · import os os.path.abspath(os.path.expanduser(os.path.expandvars(PathNameString))) Note that … WebFeb 22, 2024 · You can use glob.glob() to find files using wildcard instead of listing all the files and then go through the file list to find the required files. Also there is no xdg-open in Windows platform, use os.startfile() instead of subprocess.call().

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WebApr 4, 2024 · os.startfile () path in python with numbers Ask Question Asked 6 years ago Modified 6 years ago Viewed 9k times 1 I am working on a little project in python for work. It involves opening a file with the os.startfile () And there in lies my problem : the path to the file contains several numbers. WebSep 9, 2008 · It should be from path import Path then Path ('mydir/myfile.txt').abspath () – Frak Jun 5, 2024 at 14:51 1 There are no typos, you may have been using a different path module. The linked module uses a class named path.

WebOct 28, 2024 · try this: import os from time import sleep os.startfile ('yourFile.xclx') sleep (4) os.system ('TASKKILL /F /IM EXCEL.EXE') # os.system ('TASKKILL /F /FI "WINDOWTITLE yourtitle"') link to the source Share Follow edited Oct 28, 2024 at 7:45 answered Oct 28, 2024 at 7:30 Matiiss 5,892 2 13 29 module 'os' has no attribute 'openfile' – Soother WebYou are using splitext to determine the source filename to rename: filename_split = os.path.splitext (filename) # filename and extensionname (extension in [1]) filename_zero = filename_split [0]# ... os.rename (filename_zero, filename_zero.replace ('+','_'))

WebTo fix this, change the loop in your code to: for root, dirs, filenames in os.walk (folder): for filename in filenames: filename = os.path.join (root, filename) ... process file here. which …

WebOct 11, 2024 · Description What steps will reproduce the problem? 'conda create -n myenv spyder-kernels' in miniconda cmd 2)change default env in prefrences 3)restart console Traceback ERROR:traitlets:Failed to r...

WebNov 4, 2024 · I know it only fails because the text file is on the server and not on the computer but I have no idea how to access it. Here's my code : from os import startfile … diashow als hintergrundWebDec 7, 2024 · Python's installer doesn't define "runas" for "Python.File", but you can create it by copying the "open" command. Then you can use os.startfile to run a script elevated, as long as it doesn't require command-line arguments. If you need command-line arguments, then you'll have to use PyWin32 or ctypes to call ShellExecute [Ex] directly. – Eryk Sun citi hardware roofing price listWebMar 21, 2024 · import os, sys, subprocess def open_file (filename): if sys.platform == "win32": os.startfile (filename) else: opener = "open" if sys.platform == "darwin" else "xdg-open" subprocess.run ( [opener, filename]) If your file is just a bash script, you can replace the subprocess.run line by subprocess.run ( ["bash", filename]) Share diashow als hintergrund windows 10WebDec 19, 2024 · 1 I saw a question on here somewhat like mine, but the solution there did not work. My code is: for filename in os.scandir ('\\\\network_drive\\folder\\folder\\folder\\'): … citihardware shopeeWebJul 28, 2011 · There is no option to wait for the application to close, and no way to retrieve the application’s exit status. If you know the path of the application to open the file with, … citi hardware shop onlineWebOct 14, 2024 · import os file_name = raw_input("File Name: ") #file to be searched #cur_dir = raw_input("Search Directory: ") # Dir from where search starts can be replaced with … citi hardware san jose mindoroWebThe file itself is located in the same folder as the script file trying to open it: C:\Users\User\Desktop\Python stuff\Data.txt for simplicity, the simplest means to access the file (at least that I know of) is f=open These lines were coded as: f = open ("Data.txt", "r") and f = open ("C:/Users/User/Desktop/Python stuff/Data.txt", "r") diashow amazon fire