I've read that there is a way to open a file in python with..
os.startfile('file.exe')
is there a way to close the same file when open?
Thank you in advance!
From the os.startfile() doc:
startfile() returns as soon as the associated application is launched. There is no option to wait for the application to close, and no way to retrieve the application’s exit status.
So, basically, no, there isn't a way to close a file opened with startfile.
It isn't clear from the question is whether you want to launch a file, or to open it (for reading/writing).
If you want to launch a process, subprocess is a better candidate for running other processes and controlling them through a subshell (including killing them.)
If you want to open a file for read/write, then open() would be a good choice to start with.
As the Python Wiki says this function does the following:
Start a file with its associated application.
So the best idea would be to use os.kill to kill the application in which it is opened. The problem lies in identyfying what application is associated with the file of following extension and finding pid of exact instance which opened that file.
You have used .exe file in example which is executable file extension, so you probably misunderstood what this function does. What are you trying to accomplish? Are you sure it is the correct way of doing it?
If you really want to launch executable file, you should probably use os.system(). If you want to create a new file, write something in it and close it, look for python file operations, here are good examples: http://www.tutorialspoint.com/python/python_files_io.htm
Related
I've read that there is a way to open a file in python with..
os.startfile('file.exe')
is there a way to close the same file when open?
Thank you in advance!
From the os.startfile() doc:
startfile() returns as soon as the associated application is launched. There is no option to wait for the application to close, and no way to retrieve the application’s exit status.
So, basically, no, there isn't a way to close a file opened with startfile.
It isn't clear from the question is whether you want to launch a file, or to open it (for reading/writing).
If you want to launch a process, subprocess is a better candidate for running other processes and controlling them through a subshell (including killing them.)
If you want to open a file for read/write, then open() would be a good choice to start with.
As the Python Wiki says this function does the following:
Start a file with its associated application.
So the best idea would be to use os.kill to kill the application in which it is opened. The problem lies in identyfying what application is associated with the file of following extension and finding pid of exact instance which opened that file.
You have used .exe file in example which is executable file extension, so you probably misunderstood what this function does. What are you trying to accomplish? Are you sure it is the correct way of doing it?
If you really want to launch executable file, you should probably use os.system(). If you want to create a new file, write something in it and close it, look for python file operations, here are good examples: http://www.tutorialspoint.com/python/python_files_io.htm
I want to close an excel file after opening it. I searched this site but I cannot get a solution. Can anyone help?
import os
#TODO: close the file 'myexcel.xlsx' before writing
#writing data to 'myexcel.xlsx' here
a = os.startfile('myexcel.xlsx')#open the file for reading
Thank you very much.
This isn't possible using startfile(). According to the os.startfile documentation:
startfile() returns as soon as the associated application is launched. There is no option to wait for the application to close, and no way to retrieve the application’s exit status.
You might consider using the subprocess module. There are a load of options depending on what exactly you're trying to do at this question: How to terminate a python subprocess launched with shell=True
I'm working on a script that should check on certain system events (like opening of a file, or changing of a registry key) and start further actions depending on that. But I haven't found a clean way to get the information into my script.
I'm looking for a way to get the output of Sysinternals Process Monitor into another program. This should happen without user interaction in close to real time; so saving into a CSV/XML and than using this doesn't work.
I've checked on using the backing file, but this is in the Process Monitor PML format, which i haven't found to be documented anywhere.
Does anybody know a way how I can get the output of Process Monitor into my script?
Or an other (not too messy) way to get a real time list of opened files, registry keys etc into a python program?
Thanks!
If you want to parse stdout or a file, and your ok with a 32 bit only solution, try Dr Strace or ntstrace.
YOu could also look into ospy or another ProcMon alternative. ospy is open source, so at the very least you could look at the source code for capturing events.
Here is a list of alternates to ProcMon.
How to get open files of a subprocess?
i opened a subprocess which generate files, i want get file descritor of these files to do fsync on them
so if i have code like this:
p = subprocess.Popen([
'some_program'
])
the process p generate some files
i can get the process id of the subprocess using:
p.pid
but how can i get fd of these files to call flush and fsync() on them?
actually i find a utility called "lsof" (list open files)
but it is not installed or supported on my system, so i did not do further investigations on it, as i really need a standard way
thanks
Each process has its own table of file descriptors. If you know that a child process has a certain file open with FD 8 (which is easy enough, just take a listing of /proc/<pid>/fd), when you do fsync(8) you are sync'ing a file of your process, not the child's.
The same applies to all functions that use file descriptors: fread, fwrite, dup, close...
To get the effect of fsync, you might call sync instead.
What you could do instead is implement some kind of an RPC mechanism. For example you could add a signal handler that makes the child run fsync on all open FDs when it receives SIGUSR1.
If you want to use a packed solution, instead of going to /proc/pid/fd, an option is to use lsof of psutils
You can't fsync on behalf of another process. Also, you probably want flushing, not fsync. You can't flush on behalf of another process either. Rethink your requirements.
I'm trying to build an application that displays in a GUI the contents of a log file, written by a separate program that I call through subprocess. The application runs in Windows, and is a binary that I have no control over. Also, this application (Actel Designer if anyone cares) will write its output to a log file regardless of how I redirect the output of subprocess, so using a pipe for the output doesn't seem to be an option. The bottom line is that I seem to be forced into reading from a log file at the same time another thread may be writing to it. My question is if there is a way that I can keep the GUI's display of the log file's contents up to date in a robust way?
I've tried the following:
Naively opening the file for reading periodically while the child
process is running causes Python to crash (I'm guessing because the
child thread is writing to the file while I'm attempting to read its
contents)
Next I tried to open a file handle to the log filename before invoking the child process with GENERIC_READ, and SHARED_READ | SHARED_WRITE | SHARED_DELETE and reading back from that file. With this approach, the file appears empty
Thanks for any help you can provide - I'm not a professional programmer and I've been pulling my hair out over this for a week.
You should register for notifications on file change, the way tail -f does (you can find out what system calls it uses by executing strace tail -f logfile).
pyinotify provides a Python interface for these file change notifications.