python - automatically launch a text file created in the program

Question

I did the search but i couldn't find any help, apologies if i my question is duplicate.

i am writing the code with python 3.6 and in windows environment.in my code, i opened a text file, write the data and close the file.

self.fileName = 'file path' self.log_file = open(self.fileName, 'w') self.log_file.write('Write results') self.lof_file.close()

Instead of the user goes to file path and click to open it, i want to launch the file automatically after python save it.

how do i do that? please help

EDIT:

os.startfile(filepath=self.fileName)

command is working fine, but its opening with default program which is Notepad, how to open the file with specific program, for example, Notepad++

yes, i am creating the file as *.txt want to open it with Notepad or Notepad++ — mortuzahasan
– mortuzahasan, Commented Jun 22, 2018 at 6:34
this doesn't answer my question, how your file should be launched? in what text editor? it is OS-dependent — Azat Ibrakov
– Azat Ibrakov, Commented Jun 22, 2018 at 6:36
i am using Windows platform, instead of default program i want to open it with another program. — mortuzahasan
– mortuzahasan, Commented Jun 22, 2018 at 6:40
Have a look at the subprocess module, especially subprocess.Popen. — Jeronimo
– Jeronimo, Commented Jun 22, 2018 at 7:58

dodobhoot · Accepted Answer · 2018-06-22 06:57:58Z

If you know the command line way of doing it, you can use the os module as follows:

import os self.file = 'file path' self.log_file = open(self.fileName, 'w') self.log_file.write('Write results') self.lof_file.close() os.system('gedit <file_path>') # for ubuntu, gedit is generally present

For Windows, you can use:

import os os.startfile('C:\\Users\\RandomUser\\Documents\\test.txt')

Check this answer for more details: https://stackoverflow.com/a/15055133/9332801

Sorry to ask you, can you explain a bit please, i am using windows

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