Cannot find file path startfile python

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August undefined, 2024

WebJul 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, … WebDec 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

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WebOct 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 … WebSep 9, 2008 · import os os.path.abspath(os.path.expanduser(os.path.expandvars(PathNameString))) Note that … the pringles man

python - FileNotFoundError: [WinError 2] The system cannot find …

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 … 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 … 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... theprink.in

Problem with special characters and os.path.getsize() in Python

python - Pathname too long to open? - Stack Overflow

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 ('+','_')) 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(). the pringles spiderWebJan 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 … the prink

"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. " - Cannot find file path startfile python

Cannot find file path startfile python

python - WindowsError: [Error 2] The system cannot find the file ...

WebDec 28, 2014 · Solution: You need to set options {"enable-local-file-access": ""}. For example: pdfkit.from_string (_html, pdf_path, options= {"enable-local-file-access": ""}) – Milovan Tomašević Dec 21, 2024 at 21:44 Add a comment 2 You need set: pdfkit.from_url ('http://google.com', 'out.pdf',configuration=config) Share Improve this answer Follow WebOct 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 …

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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 … 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. WebFeb 22, 2024 · I am trying to select a random music file from a folder in Python using the windows commands: random.choice () os.listdir () os.startfile () Here is the code: import …

WebThe 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") 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

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 …

WebOct 31, 2014 · Create a batch file sam_ple.bat with the following commands and arguments cd C:\test\ start sample.exe -color Then put sam_ple.bat in the same directory as your script.py file Enter the following line of code in python to launch the exe: os.startfile ('.\sam_ple.bat') Share Improve this answer edited Nov 1, 2024 at 4:21 Michael M. 8,845 … the pringle tree in wvWebDec 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\\' the prinlaws leslieWebMar 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 theprinn.comWebJul 22, 2024 · However, looking at this question Open document with default OS application in Python, both in Windows and Mac OS you should use start instead of open for … the prink discount codeWebMar 11, 2012 · There is this one part of my code that causes trouble. I try to open a file using the os.system ('gnome-open ' + filePath) command but I can't get it to open a file … theprinkWebJul 19, 2024 · If you want to open the KML file according to its registered file association, use os.startfile (lines_kml_flyingpath). Error while converting pdf to htmls using subprocess Almtein (Mtein) July 19, 2024, 4:00pm 7 I attached below the complete code, its small. Script * import simplekml import subprocess import pandas as pd the prink giftWebJul 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 … the prink bomb