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How to program hotstrings in python like in autohotkey

I want to make hotstrings in python that converts one word when typed into another after some processing, since AHK is very limiting when it comes to determining which word to type. Right now, I am using a hotstring in ahk that runs code on the command line that runs a python script with the word that I typed as arguments. Then I use pyautogui to type the word. However, this is very slow and does not work when typing at speed. I'm looking for a way to do this all with python and without ahk, but I have not found a way to do hotstrings in python. For example, every time I type the word "test" it replaces it with "testing." Thanks for your help. I'm running the latest version of Python and Windows 10 if that is useful to anyone by the way.

(if you want to process it as each letter is typed(t,te,tes,test), you should edit your question)
I call my SymPy functions using ahk hotkeys. I register the python script as a COM server and load it using ahk.
I do not notice any latency.
you'll need pywin32, but don't download using pip install pywin32
download from https://github.com/mhammond/pywin32/releases
OR ELSE IT WON'T WORK for AutoHotkeyU64.exe, it will only work for AutoHotkeyU32.exe.
make sure to download amd64, (I downloaded pywin32-300.win-amd64-py3.8.exe)
here's why: how to register a 64bit python COM server
toUppercase COM server.py
class BasicServer:
# list of all method names exposed to COM
_public_methods_ = ["toUppercase"]
#staticmethod
def toUppercase(string):
return string.upper()
if __name__ == "__main__":
import sys
if len(sys.argv) < 2:
print("Error: need to supply arg (""--register"" or ""--unregister"")")
sys.exit(1)
else:
import win32com.server.register
import win32com.server.exception
# this server's CLSID
# NEVER copy the following ID
# Use "print(pythoncom.CreateGuid())" to make a new one.
myClsid="{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}"
# this server's (user-friendly) program ID
myProgID="Python.stringUppercaser"
import ctypes
def make_sure_is_admin():
try:
if ctypes.windll.shell32.IsUserAnAdmin():
return
except:
pass
exit("YOU MUST RUN THIS AS ADMIN")
if sys.argv[1] == "--register":
make_sure_is_admin()
import pythoncom
import os.path
realPath = os.path.realpath(__file__)
dirName = os.path.dirname(realPath)
nameOfThisFile = os.path.basename(realPath)
nameNoExt = os.path.splitext(nameOfThisFile)[0]
# stuff will be written here
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\CLSID\${myClsid}
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\CLSID\{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}
# and here
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\${myProgID}
# HKEY_LOCAL_MACHINE\SOFTWARE\Classes\Python.stringUppercaser
win32com.server.register.RegisterServer(
clsid=myClsid,
# I guess this is {fileNameNoExt}.{className}
pythonInstString=nameNoExt + ".BasicServer", #toUppercase COM server.BasicServer
progID=myProgID,
# optional description
desc="return uppercased string",
#we only want the registry key LocalServer32
#we DO NOT WANT InProcServer32: pythoncom39.dll, NO NO NO
clsctx=pythoncom.CLSCTX_LOCAL_SERVER,
#this is needed if this file isn't in PYTHONPATH: it tells regedit which directory this file is located
#this will write HKEY_LOCAL_MACHINE\SOFTWARE\Classes\CLSID\{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}\PythonCOMPath : dirName
addnPath=dirName,
)
print("Registered COM server.")
# don't use UseCommandLine(), as it will write InProcServer32: pythoncom39.dll
# win32com.server.register.UseCommandLine(BasicServer)
elif sys.argv[1] == "--unregister":
make_sure_is_admin()
print("Starting to unregister...")
win32com.server.register.UnregisterServer(myClsid, myProgID)
print("Unregistered COM server.")
else:
print("Error: arg not recognized")
you first need to register the python COM server:
first, get your own CLSID: just use a python shell.
import pythoncom
print(pythoncom.CreateGuid())
then, set myClsid to that output
to register:
python "toUppercase COM server.py" --register
to unregister:
python "toUppercase COM server.py" --unregister
hotstring python toUppercase.ahk
#NoEnv ; Recommended for performance and compatibility with future AutoHotkey releases.
#SingleInstance, force
SendMode Input ; Recommended for new scripts due to its superior speed and reliability.
SetWorkingDir %A_ScriptDir% ; Ensures a consistent starting directory.
SetBatchLines, -1
#KeyHistory 0
ListLines Off
#Persistent
#MaxThreadsPerHotkey 4
pythonComServer:=ComObjCreate("Python.stringUppercaser")
; OR
; pythonComServer:=ComObjCreate("{C70F3BF7-2947-4F87-B31E-9F5B8B13D24F}") ;use your own CLSID
; * do not wait for string to end
; C case sensitive
:*:hello world::
savedHotstring:=A_ThisHotkey
;theActualHotstring=savedHotstring[second colon:end of string]
theActualHotstring:=SubStr(savedHotstring, InStr(savedHotstring, ":",, 2) + 1)
send, % pythonComServer.toUppercase(theActualHotstring)
return
f3::Exitapp
you can test the speed of hotstring hello world, it's very fast for me.
Edit def toUppercase(string): to your liking

Executable out of script containing serial_for_url

I have developed a python script for making a serial communication to a digital pump. I now need to make an executable out of it. However even though it works perfectly well when running it with python and py2exe does produce the .exe properly when I run the executable the following error occurs:
File: pump_model.pyc in line 96 in connect_new
File: serial\__init__.pyc in line 71 in serial_for_url
ValueError: invalid URL protocol 'loop' not known
The relevant piece of my code is the following:
# New serial connection
def connect_new(self, port_name):
"""Function for configuring a new serial connection."""
try:
self.ser = serial.Serial(port = port_name,\
baudrate = 9600,\
parity = 'N',\
stopbits = 1,\
bytesize = 8,\
timeout = self.timeout_time)
except serial.SerialException:
self.ser = serial.serial_for_url('loop://',\
timeout = self.timeout_time) # This line BLOWS!
except:
print sys.exc_info()[0]
finally:
self.initialize_pump()
I should note that the application was written in OSX and was tested on Windows with the Canopy Python Distribution.

I had the exact same problem with "socket://" rather than "loop://"
I wasn't able to get the accepted answer to work however the following seems to succeed:
1) Add an explicit import of the offending urlhandler.* module
import serial
# explicit import for py2exe - to fix "socket://" url issue
import serial.urlhandler.protocol_socket
# explicit import for py2exe - to fix "loop://" url issue (OP's particular prob)
import serial.urlhandler.protocol_loop
# use serial_for_url in normal manner
self._serial = serial.serial_for_url('socket://192.168.1.99:12000')
2) Generate a setup script for py2exe (see https://pypi.python.org/pypi/py2exe/) -- I've installed py2exe to a virtualenv:
path\to\env\Scripts\python.exe -m py2exe myscript.py -W mysetup.py
3) edit mysetup.py to include option
zipfile="library.zip" # default generated value is None
(see also http://www.py2exe.org/index.cgi/ListOfOptions)
3) build it:
path\to\env\Scripts\python.exe mysetup.py py2exe
4) run it
dist\myscript.exe

Found it!
It seems that for some reason the 'loop://' arguement can't be recognised after the .exe production.
I figured out by studying the pyserial/init.py script that when issuing the command serial.serial_for_url(‘loop://') you essentially call:
sys.modules['serial.urlhandler.protocol_loop’].Serial(“loop://“)
So you have to first import the serial.urlhandler.protocol_loop
and then issue that command in place of the one malfunctioning.
So you can now type:
__import__('serial.urlhandler.protocol_loop')
sys.modules[‘serial.urlhandler.protocol_loop’].Serial("loop://")
After this minor workaround it worked fine.

Unable to identify the host : Fabric

I m trying to use fabric module through simple python module
remoteExc.py
from fabric.api import *
def clone_repo(IPADDRESS,USER,fPath,git_url):
env.hosts_string = IPADDRESS
env.user = USER
env.key_filename = fPath
env.disable_known_hosts = 'True'
run('git clone %s' % (git_url))
mainFile.py
from remoteExc import clone_repo
clone_repo(ipAddress,user,fPath,git_url)
When i execute it says
python mainfile.py
No hosts found. Please specify (single) host string for connection:
Please enlight me where i make a mistake

Typo. env.host_string = IPADDRESS - you've got an env.hosts_string instead.
Also, generally you run fabric via fab - unless you're trying to do something fairly non-standard, be aware that running it via python probably isn't what you want to do. See the Fabric docs for a pretty good intro.
http://docs.fabfile.org/en/1.7/tutorial.html

Easy way to suppress output of fabric run?

I am running a command on the remote machine:
remote_output = run('mysqldump --no-data --user=username --password={0} database'.format(password))
I would like to capture the output, but not have it all printed to the screen. What's the easiest way to do this?

It sounds like Managing output section is what you're looking for.
To hide the output from the console, try something like this:
from __future__ import with_statement
from fabric.api import hide, run, get
with hide('output'):
run('mysqldump --no-data test | tee test.create_table')
get('~/test.create_table', '~/test.create_table')
Belows is the sample results:
No hosts found. Please specify (single) host string for connection: 192.168.6.142
[192.168.6.142] run: mysqldump --no-data test | tee test.create_table
[192.168.6.142] download: /home/quanta/test.create_table <- /home/quanta/test.create_table

Try this if you want to hide everything from log and avoid fabric throwing exceptions when command fails:
from __future__ import with_statement
from fabric.api import env,run,hide,settings
env.host_string = 'username#servernameorip'
env.key_filename = '/path/to/key.pem'
def exec_remote_cmd(cmd):
with hide('output','running','warnings'), settings(warn_only=True):
return run(cmd)
After that, you can check commands result as shown in this example:
cmd_list = ['ls', 'lss']
for cmd in cmd_list:
result = exec_remote_cmd(cmd)
if result.succeeded:
sys.stdout.write('\n* Command succeeded: '+cmd+'\n')
sys.stdout.write(result+"\n")
else:
sys.stdout.write('\n* Command failed: '+cmd+'\n')
sys.stdout.write(result+"\n")
This will be the console output of the program (observe that there aren't log messages from fabric):
* Command succeeded: ls
Desktop espaiorgcats.sql Pictures Public Videos
Documents examples.desktop projectes scripts
Downloads Music prueba Templates
* Command failed: lss
/bin/bash: lss: command not found

For fabric==2.4.0 you can hide output using the following logic
conn = Connection(host="your-host", user="your-user")
result = conn.run('your_command', hide=True)
result.stdout.strip() # here you can get the output

As other answers allude, fabric.api doesn't exist anymore (as of writing, fabric==2.5.0) 8 years after the question. However the next most recent answer here implies providing hide=True to every .run() call is the only/accepted way to do it.
Not being satisfied I went digging for a reasonable equivalent to a context where I can specify it only once. It feels like there should still be a way using an invoke.context.Context but I didn't want to spend any longer on this, and the easiest way I could find was using invoke.config.Config, which we can access via fabric.config.Config without needing any additional imports.
>>> import fabric
>>> c = fabric.Connection(
... "foo.example.com",
... config=fabric.config.Config(overrides={"run": {"hide": True}}),
... )
>>> result = c.run("hostname")
>>> result.stdout.strip()
'foo.example.com'

As of Fabric 2.6.0 hide argument to run is not available.
Expanding on suggestions by #cfillol and #samuel-harmer, using a fabric.Config may be a simpler approach:
>>> import fabric
>>> conf = fabric.Config()
>>> conf.run.hide = True
>>> conf.run.warn = True
>>> c = fabric.Connection(
... "foo.example.com",
... config=conf
... )
>>> result = c.run("hostname")
This way no command output is printed and no exception is thrown on command failure.

As Samuel Harmer also pointed out in his answer, it is possible to manage output of the run command at the connection level.
As of version 2.7.1:
from fabric import Config, Connection
connection = Connection(
host,
config = Config(overrides = {
"run": { "hide": "stdout" }
}),
...
)

retrieving current URL from FireFox with python

I want to know what is the current url of active tab in running firefox instance from python module. Does FireFox have any API for this and does python know to work with it?

The most convenient way maybe insatll a firefox extension to open up a tcp service, then you can exchange info with firefox.
mozrepl can set up a telnet service, you can call js-like command to get info.
With telnetscript (http: //code.activestate.com/recipes/152043/), you can write:
import telnetscript
script = """rve
w content.location.href;
ru repl>
w repl.quit()
cl
"""
conn = telnetscript.telnetscript( '127.0.0.1', {}, 4242 )
ret = conn.RunScript( script.split( '\n' )).split( '\n' )
print ret[-2][6:]

If on windows you can use win32com
import win32clipboard
import win32com.client
shell = win32com.client.Dispatch("WScript.Shell")
shell.AppActivate('Some Application Title')
Then use shell.SendKeys to do a ctrl+l and a ctrl+c
Then read the string in the clipboard.
It's horkey though it will work, alternatly you can use something like AutoIt an compile the code to an exe that you can work with.
Hope this helps.