How to run an AppleScript from within a Python script?
The questions says it all..
(On a Mac obviously)
this nice article suggests the simple solution
cmd = """osascript -e 'tell app "Finder" to sleep'"""
def stupidtrick():
os.system(cmd)
though today you'd use the subprocess module instead of os.system, of course.
Be sure to also check page 2 of the article for many more info and options, including appscript.
A subprocess version which allows running an original apple script as-is, without having to escape quotes and other characters which can be tricky. It is a simplified version of the script found here which also does parametrization and proper escaping (Python 2.x).
import subprocess
script = '''tell application "System Events"
activate
display dialog "Hello Cocoa!" with title "Sample Cocoa Dialog" default button 2
end tell
'''
proc = subprocess.Popen(['osascript', '-'],
stdin=subprocess.PIPE,
stdout=subprocess.PIPE)
stdout_output = proc.communicate(script)[0]
print stdout_output
NOTE: If you need to execute more than one script with the same Popen instance then you'll need to write explicitly with proc.stdin.write(script) and read with proc.stdout.read() because communicate() will close the input and output streams.
I got the Output folks... Here it's following:
import subprocess
import sys
for i in range(int(sys.argv[1])):
ip = str(sys.argv[2])
username = str(sys.argv[3])
pwd = str(sys.argv[4])
script = '''tell application "Terminal"
activate
do script with command "cd Desktop && python test_switch.py {ip} {username} {pwd}"
delay 15
end tell
'''
proc = subprocess.Popen(['osascript', '-'],
stdin=subprocess.PIPE,
stdout=subprocess.PIPE)
stdout_output = proc.communicate(script.format(ip=ip, username=username, pwd=pwd))[0]
I was pretty frustrated at the lack of detail in Apple's own documentation regarding how to do this AND to also pass in arguments. I had to send the desired arg (in this case a zoom id) as a string otherwise the argument didn't come through to the applescript app
Here's my code running from python:
f = script if os.path.exists(script) else _tempfile()
if not os.path.exists(script):
open(f,'w').write(script)
args = ["osascript", f, str(zoom_id)]
kwargs = {'stdout':open(os.devnull, 'wb'),'stderr':open(os.devnull, 'wb')}
#kwargs.update(params)
proc = subprocess.Popen(args,**kwargs)
and here is my applescript:
on run argv
set zoom_id to 0
zoom_id = item 1 in argv
tell application "zoom.us"
--do stuff
end tell
end run
Related
Python 3.10.6
Windows 10
I have a python function that executes a DXL script using subsystem.run() or os.system() (whichever works best I guess). The problem is that when I run a custom command using python it does not work, but when I paste the same command in the command prompt, it works. I should also clarify that command prompt is not the ms store windows terminal (cannot run ibm doors commands there for some reason). It is the OG prompt
I need to use both python and IBM Doors for the solution.
Here is a summer version of my code (Obviously, the access values are not real):
#staticmethod
def run_dxl_importRTF():
dquotes = chr(0x22) # ASCII --> "
module_name = "TEST_TEMP"
script_path = "importRTF.dxl"
script_do_nothing_path = "doNothing.dxl"
user = "user"
password = "pass"
database_config = "11111#11.11.1111.0"
doors_path = dquotes + r"C:\Program Files\IBM\Rational\DOORS\9.7\bin\doors.exe" + dquotes
file_name = "LIBC_String.rtf"
# Based On:
# "C:\Program Files\IBM\Rational\DOORS\9.7\\bin\doors.exe" -dxl "string pModuleName = \"%~1\";string pFilename = \"%~2\";#include <importRTF.dxl>" -f "%TEMP%" -b "doNothing.dxl" -d 11111#11.11.1111.0 -user USER -password PASSWORD
script_arguments = f"{dquotes}string pModuleName=\{dquotes}{module_name}\{dquotes};string pFileName=\{dquotes}{file_name}\{dquotes};#include <{script_path}>{dquotes}"
command = [doors_path, "-dxl", script_arguments, "-f", "%TEMP%", "-b", script_do_nothing_path, '-d', database_config, '-user', user, '-password', password]
res = subprocess.run(command, shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE, text=True)
print(f"COMMAND:\n{' '.join(res.args)}")
print(f"STDERR: {repr(res.stderr)}")
print(f'STDOUT: {res.stdout}')
print(f'RETURN CODE: {res.returncode}')
return
PYTHON SCRIPT OUTPUT:
COMMAND:
"C:\Program Files\IBM\Rational\DOORS\9.7\bin\doors.exe" -dxl "string pModuleName=\"TEST_TEMP\";string pFileName=\"LIBC_String.rtf\";#include <importRTF.dxl>" -f %TEMP% -b doNothing.dxl -d 11111#11.11.1111.0 -user USER_TEMP -password PASS_TEMP
STDERR: 'The system cannot find the path specified.\n'
STDOUT:
RETURN CODE: 1
When I run the same command in the command prompt, it works (dxl script is compiled).
I identified the problem which is the script_argument variable. Meaning that, when I try to just enter the IBM Doors server without compiling a DXL script, it works on python and the command prompt.
The python script needs to be dynamic meaning that all of the initial declared variables can change value and have a path string in it. I am also trying to avoid .bat files. They also did not work with dynamic path values
Thanks for your time
I tried:
Changing CurrentDirectory (cwd) to IBM Doors
os.system()
Multiple workarounds
Tried IBM Doors path without double quotes (it doesnt work because of the whitespaces)
.bat files
When calling subprocess.run with a command list and shell=True, python will expand the command list to a string, adding more quoting along the way. The details are OS dependent (on Windows, you always have to expand the list to a command) but you can see the result via the subprocess.list2cmdline() function.
Your problem is these extra escapes. Instead of using a list, build a shell command string that already contains the escaping you want. You can also use ' for quoting strings so that internal " needed for shell quoting can be entered literally.
Putting it all together (and likely messing something up here), you would get
#staticmethod
def run_dxl_importRTF():
module_name = "TEST_TEMP"
script_path = "importRTF.dxl"
script_do_nothing_path = "doNothing.dxl"
user = "user"
password = "pass"
database_config = "11111#11.11.1111.0"
doors_path = r"C:\Program Files\IBM\Rational\DOORS\9.7\bin\doors.exe"
file_name = "LIBC_String.rtf"
script_arguments = (rf'string pModuleName=\"{module_name}\";'
'string pFileName=\"{file_name}\";'
'#include <{script_path}>')
command = (f'"{doors_path}" -dxl "{script_arguments}" -f "%TEMP%"'
' -b "{script_do_nothing_path}" -d {database_config}'
' -user {user} -password {pass}')
res = subprocess.run(command, shell=True, stdout=subprocess.PIPE, stderr=subprocess.PIPE, text=True)
print(f"COMMAND:\n{' '.join(res.args)}")
print(f"STDERR: {repr(res.stderr)}")
print(f'STDOUT: {res.stdout}')
print(f'RETURN CODE: {res.returncode}')
I have a cmd file "file.cmd" containing 100s of lines of command.
Example
pandoc --extract-media -f docx -t gfm "sample1.docx" -o "sample1.md"
pandoc --extract-media -f docx -t gfm "sample2.docx" -o "sample2.md"
pandoc --extract-media -f docx -t gfm "sample3.docx" -o "sample3.md"
I am trying to run these commands using a script so that I don't have to go to a file and click on it.
This is my code, and it results in no output:
file1 = open('example.cmd', 'r')
Lines = file1.readlines()
# print(Lines)
for i in Lines:
print(i)
os.system(i)
You don't need to read the cmd file line by line. you can simply try the following:
import os
os.system('myfile.cmd')
or using the subprocess module:
import subprocess
p = subprocess.Popen(['myfile.cmd'], shell = True, close_fds = True)
stdout, stderr = proc.communicate()
Example:
myfile.cmd:
#ECHO OFF
ECHO Grettings From Python!
PAUSE
script.py:
import os
os.system('myfile.cmd')
The cmd will open with:
Greetings From Python!
Press any key to continue ...
You can debug the issue by knowing the return exit code by:
import os
return_code=os.system('myfile.cmd')
assert return_code == 0 #asserts that the return code is 0 indicating success!
Note: os.system works by calling system() in C can only take up to 65533 arguments after a command (so it is a 16 bit issue). Giving one more argument will result in the return code 32512 (which implies the exit code 127).
The subprocess module provides more powerful facilities for spawning new processes and retrieving their results; using that module is preferable to using this function (os.system('command')).
since it is a command file (cmd), and only the shell can run it, then shell argument must set to be true. since you are setting the shell argument to true, the command needs to be string form and not a list.
use the Popen method for spawn a new process and the communicte for waiting on that process (you can time it out as well). if you whish to communicate with the child process, provide the PIPES (see mu example, but you dont have to!)
the code below for python 3.3 and beyond
import subprocess
try:
proc=subprocess.Popen('myfile.cmd', shell=True, stderr=subprocess.PIPE, stdout=subprocess.PIPE)
outs, errs = proc.communicate(timeout=15) #timing out the execution, just if you want, you dont have to!
except TimeoutExpired:
proc.kill()
outs, errs = proc.communicate()
for older python versions
proc = subprocess.Popen('myfile.cmd', shell=True)
t=10
while proc.poll() is None and t >= 0:
print('Still waiting')
time.sleep(1)
t -= 1
proc.kill()
In both cases (python versions) if you dont need the timeout feature and you dont need to interact with the child process, then just, use:
proc = subprocess.Popen('myfile.cmd', shell=True)
proc.communicate()
I have a program with a GUI that runs an external program through a Popen call:
p = subprocess.Popen("<commands>" , stdout=subprocess.PIPE , stderr=subprocess.PIPE , cwd=os.getcwd())
p.communicate()
But a console pops up, regardless of what I do (I've also tried passing it NUL for the file handle). Is there any way to do that without getting the binary I call to free its console?
From here:
import subprocess
def launchWithoutConsole(command, args):
"""Launches 'command' windowless and waits until finished"""
startupinfo = subprocess.STARTUPINFO()
startupinfo.dwFlags |= subprocess.STARTF_USESHOWWINDOW
return subprocess.Popen([command] + args, startupinfo=startupinfo).wait()
if __name__ == "__main__":
# test with "pythonw.exe"
launchWithoutConsole("d:\\bin\\gzip.exe", ["-d", "myfile.gz"])
Note that sometimes suppressing the console makes subprocess calls fail with "Error 6: invalid handle". A quick fix is to redirect stdin, as explained here: Python running as Windows Service: OSError: [WinError 6] The handle is invalid
just do subprocess.Popen([command], shell=True)
According to Python 2.7 documentation and Python 3.7 documentation, you can influence how Popen creates the process by setting creationflags. In particular, the CREATE_NO_WINDOW flag would be useful to you.
variable = subprocess.Popen(
"CMD COMMAND",
stdout = subprocess.PIPE, creationflags = subprocess.CREATE_NO_WINDOW
)
This works nicely in the win32api. The other solutions were not working for me.
import win32api
chrome = "\"C:\\Program Files (x86)\\Google\\Chrome\\Application\\chrome.exe\""
args = "https://stackoverflow.com"
win32api.WinExec(chrome + " " + args)
You might be able to just do subprocess.Popen([command], shell=False).
That's what I use anyways. Saves you all the nonsense of setting flags and whatnot.
Once named as a .pyw or run with pythonw it shouldn't open a console.
I created a web interface in Grails 3 where you can start different pipelines written in python via a web environment. I have created a simple form with a start button. The idea is now that when you press the start button the python pipeline is started. I can't figure it out I have tried several things a example is:
def cmd = "python amplicon_pipeline.py -i 'inputdir' -o 'outputdir' -a 'amplicon'"
def proc = cmd.execute()
proc.waitFor()
But nothing happens.
How can I get an external python script to start working when you press the start button?
Mention the full path to python and the script file.
Ex:
def cmd = ["/usr/bin/python", "/home/rm93/Documents/project_rivm/RIVM_amplicon_pipeline/amplicon_pipeline.py", "-i", "/home/rm93/Documents/Git/BIGC_test_upload/amplicon_pipeline/18/upload/", "-o", "/home/rm93/Documents/Git/BIGC_test_upload/amplicon_pipeline/18/output/", "-a", "16sv4"]
def proc = cmd.execute()
proc.waitFor()
println proc.text
I want to make a python script that will convert a TEX file to PDF and then open the output file with my document viewer.
I first tried the following:
import subprocess
subprocess.Popen(['xelatex', '--output-directory=Alunos/', 'Alunos/' + aluno + '_pratica.tex'], shell=False, stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
subprocess.Popen(['gnome-open', 'Alunos/'+aluno+'_pratica.pdf'], shell=False)
This way, the conversion from TEX to PDF works all right, but, as it takes some time, the second command (open file with Document Viewer) is executed before the output file is created.
So, I tried do make the program wait some seconds before executing the second command. Here's what I've done:
import subprocess
import time
subprocess.Popen(['xelatex', '--output-directory=Alunos/', 'Alunos/' + aluno + '_pratica.tex'], shell=False, stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
time.sleep(10)
subprocess.Popen(['gnome-open', 'Alunos/'+aluno+'_pratica.pdf'], shell=False)
But, when I do so, the output PDF file is not created. I can't understand why. The only change was the time.sleep command. Why does it affect the Popen process?
Could anyone give me some help?
EDIT:
I've followed the advice from Faust and Paulo Bu and in both cases the result is the same.
When I run this command...
subprocess.call('xelatex --output-directory=Alunos/ Alunos/{}_pratica.tex'.format(aluno), shell=True)
... or this...
p = subprocess.Popen(['xelatex', '--output-directory=Alunos/', 'Alunos/' + aluno + '_pratica.tex'], shell=False, stdin=subprocess.PIPE, stdout=subprocess.PIPE, stderr=subprocess.PIPE)
p.wait()
...the Xelatex program is run but doesn't make the conversion.
Strangely, when I run the command directly in the shell...
$ xelatex --output-directory=Alunos/ Alunos/name_pratica.tex
... the conversion works perfectly.
Here's what I get when I run the subprocess.call() command:
$ python my_file.py
Enter name:
name
This is XeTeX, Version 3.1415926-2.4-0.9998 (TeX Live 2012/Debian)
restricted \write18 enabled.
entering extended mode
(./Alunos/name_pratica.tex
LaTeX2e <2011/06/27>
Babel <v3.8m> and hyphenation patterns for english, dumylang, nohyphenation, loaded.
)
*
When I write the command directly in the shell, the output is the same, but it followed automatically by the conversion.
Does anyone know why it happens this way?
PS: sorry for the bad formating. I don't know how to post the shell output properly.
If you need to wait the termination of the program and you are not interested in its output you should use subprocess.call
import subprocess
subprocess.call(['xelatex', '--output-directory=Alunos/', 'Alunos/{}_pratica.tex'.format(aluno)])
subprocess.call([('gnome-open', 'Alunos/{}_pratica.pdf'.format(aluno)])
EDIT:
Also it is generally a good thing to use English when you have to name variables or functions.
If xelatex command works in a shell but fails when you call it from Python then xelatex might be blocked on output in your Python code. You do not read the pipes despite setting stdout/stderr to PIPE. On my machine the pipe buffer is 64KB therefore if xelatex output size is less then it should not block.
You could redirect the output to os.devnull instead:
import os
import webbrowser
from subprocess import STDOUT, check_call
try:
from subprocess import DEVNULL # py3k
except ImportError:
DEVNULL = open(os.devnull, 'w+b')
basename = aluno + '_pratica'
output_dir = 'Alunos'
root = os.path.join(output_dir, basename)
check_call(['xelatex', '--output-directory', output_dir, root+'.tex'],
stdin=DEVNULL, stdout=DEVNULL, stderr=STDOUT)
webbrowser.open(root+'.pdf')
check_call is used to wait for xelatex and raise an exception on error.