Cannot find file path startfile python
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 … WebMar 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
Cannot find file path startfile python
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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\\'
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 … WebJul 31, 2024 · 2 Answers Sorted by: 1 You can try os.system with open. For example: os.system ("open Untitled.pdf") This will open the file Untitled.pdf with the default PDF application ('Preview', in my case). Share Improve this answer Follow answered Jul 31, 2024 at 15:33 Roy2012 11.5k 2 22 35 it's recommended to use subprocess over …
WebJan 2, 2024 · 1. 1) Try this first. import openpyxl wb = openpyxl.load_workbook ('C:\Users\ my file location here.xlsx') type (wb) 2) else put your .py file in the same directory where … 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.
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 how alkalinity of water can be determinedWebDec 30, 2024 · import os def find_files (filename, search_path): for root, dir, files in os.walk (search_path): if filename in files: return os.path.join (root, filename) find = find_files ("Among Us.exe","D:") os.startfile (find) Share Improve this answer Follow edited Dec 30, 2024 at 4:09 answered Dec 30, 2024 at 4:04 Delupara 349 1 4 9 Add a comment how allen keys are madeWebNov 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 … how all corporations are linked at the topWebJan 30, 2024 · In python, we use the method os.startfile () to start or open the file. To use this method, we pass a parameter to the method in string data type, which shows a valid … how all 50 states got their namesWebOct 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 … how allah akbar a discord serverWebOct 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 … how all can a hourse survive the desertWebThe 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") how a llc and a corporation are different