Well I searched a lot and found different ways to open program in python,
For example:-
import os
os.startfile(path) # I have to give a whole path that is not possible to give a full path for every program/software in my case.
The second one that I'm currently using
import os
os.system(fileName+'.exe')
In second example problem is:-
If I want to open calculator so its .exe file name is calc.exe and this happen for any other programs too (And i dont know about all the .exe file names of every program).
And assume If I wrote every program name hard coded so, what if user installed any new program. (my program wont able to open that program?)
If there is no other way to open programs in python so Is that possible to get the list of all install program in user's computer.
and there .exe file names (like:- calculator is calc.exe you got the point).
If you want to take a look at code
Note: I want generic solution.
There's always:
from subprocess import call
call(["calc.exe"])
This should allow you to use a dict or list or set to hold your program names and call them at will. This is covered also in this answer by David Cournapeau and chobok.
You can try with os.walk :
import os
exe_list=[]
for root, dirs, files in os.walk("."):
#print (dirs)
for j in dirs:
for i in files:
if i.endswith('.exe'):
#p=os.getcwd()+'/'+j+'/'+i
p=root+'/'+j+'/'+i
#print(p)
exe_list.append(p)
for i in exe_list :
print('index : {} file :{}'.format(exe_list.index(i),i.split('/')[-1]))
ip=int(input('Enter index of file :'))
print('executing {}...'.format(exe_list[ip]))
os.system(exe_list[ip])
os.getcwd()+'/'+i prepends the path of file to the exe file starting from root.
exe_list.index(i),i.split('/')[-1] fetches just the filename.exe
exe_list stores the whole path of an exe file at each index
Can be done with winapps
First install winapps by typing:
pip install winapps
After that use the library:
# This will give you list of installed applications along with some information
import winapps
for app in winapps.list_installed():
print(app)
If you want to search for an app you can simple do:
application = 'chrome'
for app in winapps.search_installed(application):
print(app)
Related
I'm currently working on a program that uses databases and all that kind of stuff for a friend of mine. Now while doing my basic planning of what I will need to do, I thought to myself how will I get the program to connect to the right MS Access database if I hard code the path to my own personal computer? Because ill be using pyinstaller to package the program for him.
Do I need to build the directory in a dynamic way that uses the path of the current script, and then some string manipulation of that path to get the right file?
i.e.
import os
file_path = os.path.realpath(__file__)
# Using prior knowledge knowing the file name length is 7
file_path = file_path[:-7]
#appending the path string to point to the database
file_path = file_path + "/databases/test.accdb"
Or will the program still someway find its way to that database?
You can use a: os.walk(...) function to locate specified name file or every database file in specified directories. Next user can choose which file is correct
first post here so sorry if it's hard to understand. Is it possible to shorten the directory in python to the location of the .py file?. For example, if I wanted to grab something from the directory "C:\Users\Person\Desktop\Code\Data\test.txt", and if the .py was located in the Code folder, could I shorten it to "\data\test.txt". I'm new to python so sorry if this is something really basic and I just didn't understand it correctly.
I forgot to add i plan to use this with multiple files, for example: "\data\test.txt" and \data\test2.txt
import os
CUR_FILE = os.path.abspath(__file__)
TARGET_FILE = "./data/test.txt"
print(os.path.join(CUR_FILE, TARGET_FILE))
With this, you can move around your Code directory anywhere and not have to worry about getting the full path to the file.
Also, you can run the script from anywhere and it will work (you don't have to move to Code's location to run the script.
You can import os and get current working directory ,this will give you the location of python file and then you can add the location of folder data and the file stored in that ,code is given below
import os
path=os.getcwd()
full_path1=path+"\data\test.txt"
full_path2=path+"\data\test2.txt"
print(full_path1)
print(full_path2)
I think this will work for your case and if it doesn't work then add a comment
I've got a task to do that is crushing my head. I have five .py documents and I want to make a menu in another .py so I can run any of them by introducing a string inside an input() but don't really see the way to do that and I don't know if there is somehow I can.
I have tried import every file to the 6th file but I don't even know how to start.
I would like it just to be seen as simple as it can sound, but yet I find it really hard.
If you just want to run them, then try this:-
import os
file_path = input("Enter the path of your file = ")
os.system(file_path)
If the file that you are trying to execute is not in the current
directory, i.e. doesn't exist in the same folder as the currently
executing python file, then you have to provide it's full path.
Path Format:-. C:\Users\lmYoona\OneDrive\Desktop\example.py
If the python file you are trying to execute is in the same directory as
the currently executing python file, then abstract name will also
work
Path Format:- example.py
P.S.:- I would only recommend this method if all you want is just to execute the other python file, rather then importing stuff from it.
Probably a simple query.. But basically, I have data in directory "/foo/bar/foobar.txt"
and I am working in directory "/some/path/read_foobar.py"..
Now I want to read the file "foobar.txt" but rather than giving full path, I thought of adding /foo/bar/ to the path..
So, added the following at the start of read_foobar.py
import sys
sys.path.append("/foo/bar")
But when I try to read open("foobar.txt","r"), it is not able to find the file?
how do I do this?
Thanks
You can do it like this:
import os
os.chdir('/foo/bar')
f = open('foobar.txt', 'r')
sys.path is used to set the path used to look for python modules. Short of you writing some helper function that has a list of directories to search in when opening a file, I don't believe there is a standard module that provides this functionality.
From what I gathered from here and some quick tests, appending a path to sys.path will make python search in that path when you import a file/module, but not when open-ing it. Let's say we have a file called foo.py in /foo/bar/
import sys
sys.path.append("/foo/bar/")
try:
f = open('foo.py', 'r')
except:
print('this did not work') # this will print
import foo # no problems here
Unfortunately you can't. The PATH environment variable is only used by the operating system to search for executable files, and python uses it (along with the environment variable PYTHONPATH) to search for python modules to import.
You may want to consider setting a symbolic link to that file from your current working directory
ln -s /foo/bar/foobar.txt /some/path/foobar.text
Hi I am currently a beginner to the python language, it is also my first language too. I need some help I am finding it difficult to know what to use to generate permanent directories sub directories and files, for eg; I want the path to generate whatever path i enter if the directories etc. don't exist, i want them created, so I enter C:\user\python\directory\sub-directory\file, then i cant workout what i should import to do the following job.
I am using Python 3.2, any advice?
You can do:
import os
os.makedirs('a/b/c', exist_ok=True)
http://docs.python.org/py3k/library/os.html
f = open("c:\file\path","w")
f.write("content of file")
First, you open the file, storing it in variable f.
You then write to it, using f.write()
Python will create the file and path if it does not exist, I think. (I am sure I've done this before, but I can't remember)
When you have finished using the file, you should use
f.close()
to close the file safely.