Cannot find file path startfile python

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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 … 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, …

windows - Python os.startfile Can

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 … WebDec 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 many centimeters are in a gram

How to get an absolute file path in Python - Stack Overflow

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 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 … 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 how many centimeters are in 6 m

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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 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 … high school debut manga read onlineWebYou 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 … how many centimeters are in 9 kilometers

"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 … " - Cannot find file path startfile python

Cannot find file path startfile python

Python Unittest: Open and wait until the program is closed

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. 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 ('+','_'))

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

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 … 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\\'

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 … 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

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.

WebJul 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 … high school debut manga freeWebFeb 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 … high school debut ch 1WebOct 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 high school debutWebOct 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 … high school debut onlineWebThe 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") high school debate worksheetWebApr 10, 2015 · Use GetShortPathName from kernel32.dll and access the file in this way. That is nice, but I cannot use it, since I need to use the paths in a way shutil.rmtree … how many centimeters are in a inWebJan 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 … high school decal codes