I have a python script to open maya file and I want to set the project before opening the file ?
I try to find a file or an environement variable which is defining the current project path but I don't if it exists, do you know how I can do it ?
So you have a python script that opens the maya file. The file that sets the project is workspace.mel
import maya.cmds as cmds
cmds.workspace("workspace_directory", openWorkspace=True)
the cmds.workspace() command is what sets your project. "workspace_directory" is the path to where the workspace.mel command lives. You can use os.walk with the topdown=False flag in conjunction with an fnmatch to walk up directories from the desired scene file and search until it finds the workspace.mel file, and store that directory to a string. Then pass that into the "workspace_directory" argument I outline above!
https://knowledge.autodesk.com/support/maya/learn-explore/caas/CloudHelp/cloudhelp/2015/ENU/Maya/files/Interface-overview-Start-Maya-from-the-command-line--htm.html
You have all files describe in the doc. You may be interrested into :
-command [mel command]
-proj [dir]
-script [file]
Related
[Introduction]
Hi! I'm working on a python script to automate installation of UWP-Apps, it's been a long time i'm not touching Python; until this day. The script uses Depedencies inside the script directory, i've looking up on my older scripts and found this specific code.
os.chdir(os.path.dirname(sys.argv[0]))
[Problematic]
However, using the above code doesn't work on my current script but it's working fine on older scripts. When using above, it shows:
OSError: [WinError 123] The filename, directory name, or volume label syntax is incorrect: ''
Already looking up on Internet about this topic; but most of them was talking about running the script from outer/different directory that leads me to dead end.
Any helps is appreciated :)
The easiest answer is probably to change your working directory, then call the .py file from where it is:
cd path/to/python/file && python ../.py
Of course you might find it even easier to write a script that does it all for you, like so:
Save this as runPython.sh in the directory where you're running the python script from, is:
#!/bin/sh
cd path/to/python/file
python ../script.py
Make it executable (for yourself):
chmod +x ./runPython.sh
Then you can simply enter your directory and run it:
./runPython.sh
If you want to only make changes to the python script:
mydir = os.getcwd() # would be the folder where you're running the file
mydir_tmp = "path/to/python/file" # get the path
mydir_new = os.chdir(mydir_tmp) # change the current working directory
mydir = os.getcwd()
The reason you got an error was because sys.argv[0] is the name of the file itself, instead you can pass the directory as a string and use sys.argv[1] instead.
import os
from os.path import abspath, dirname
os.chdir(dirname(abspath(__file__)))
You can use dirname(abspath(__file__))) to get the parent directory of the python script and os.chdir into it to make the script run in the directory where it is located.
Basically, I want to call a python program from anywhere using shell so I added a shebang and copied it to /usr/local/bin with executable permission. The python program takes a command line argument which is the relative path of an input file.
I am stuck here, I have no idea what to do so that I can obtain the absolute path of the shell. I am assuming once I get the absolute path somehow, I can use sys.argv[1] to get the entered relative path of the file(which I will append to the absolute path of shell working directory) but please do correct me it won't work.
You can Print Working Directory via os.environ['PWD']. Content of your_script.py:
#!/usr/bin/python3.5
import os
print(os.environ['PWD'])
Usage:
sanyash#sanyash-ub16:/etc/nginx$ your_script.py
/etc/nginx
sanyash#sanyash-ub16:/etc/nginx$
I have some problem with my WIndows CMD.
Some time I need to open python file using CMD command. And I write: 'C:\Program Files\Python X.X\python.exe file.py' but have error: 'C:\Program' isn't system command (maybe not the same, I have another OS language).
With different methods I have different errors but can't open python file.
Examples:
(Picture) translate: can't find 'C:\Program'...
(Picture) another example when I trying to write python directory first and then start python file, but it can't find python file.
Thanks for helping me.
There seems to be 2 different problems here.
Windows does not recognise spaces in directory or file names on the command line, so you need to put the directory insied "" .
i.e. "C:\Program Files\Python 3.4\python.exe"
In your second picture, suggests that run.py does not exist in the current directory. Change Directory to where the run.py file is before running that command.
First of all go to the directory where your python file is located ... like:
cd "c:\users\someone\documents\..."
On your pictures you are trying to run python file located in system32 folder but i guess it is not located there so move where the file is with that cd command
Then as Martin says the problem with path of python.exe is the space between words. To solve put the path into quotation marks.
But u can add python to system path and insted of writing full path u can write only
python file.py
How to add python to path see here https://superuser.com/questions/143119/how-to-add-python-to-the-windows-path
My Crontab -l
# m h dom mon dow command
SHELL=/bin/bash
PATH=/usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games
00 8,20 * * * python /home/tomi/amaer/controller.py >>/tmp/out.txt 2>&1
My controller.py has config file settings.cfg also it uses other script in the folder it's located (I chmoded only controller.py)
The error
1;31mIOError^[[0m: [Errno 2] No such file or directory: 'settings.cfg'
I have no idea how to fix this? Please help me?
Edit: The part that read the config file
def main():
config=ConfigParser.ConfigParser()
config.readfp(open("settings.cfg"),"r")
As I initially wrote in my comment, this is because you are using relative path to the current working directory. However, that is not going to be the same when running all this via the cron executable rather than the python interpreter directly via the shebang.
Your current code would look for the "settings.cfg" in the current working directory which is where the cron executable resides, and not your script. Hence, you would need to change your code logic to using absolute paths by the help of the "os" built-in standard module.
Try to following line:
import os
...
def main():
config = ConfigParser.ConfigParser()
scriptDirectory = os.path.dirname(os.path.realpath(__file__))
settingsFilePath = os.path.join(scriptDirectory, "settings.cfg")
config.readfp(open(settingsFilePath,"r"))
This will get your the path of your script and then appends the "settings.cfg" with the appropriate dir separator for your operating system which is Linux in this particular case.
If the location of the config file changes any time in the future, you could use the argparse module for processing a command line argument to handle the config location properly, or even without it simply just using the first argument after the script name like sys.argv[1].
Your code is looking for settings.cfg in its current working directory.
This working directory will not be the same when cron executes the job, hence the error
You have two "easy" solutions:
Use an absolute path to the config file in your script (/home/tomi/amaer/config.cfg)
CD to the appropriate directory first in your crontab (cd /home/tomi/amaer/ && python /home/tomi/amaer/controller.py)
The "right" solution, though, would be to pass your script a parameter (or environment variable) that tells it where to look for the config file.
It's not exactly good practice to assume your config file will always be lying just next to your script.
You might want to have alook at this question: https://unix.stackexchange.com/questions/38951/what-is-the-working-directory-when-cron-executes-a-job
I would like to do a very simple thing but I am quite lost.
I am using a program called Blender and I want to write a script in python which open a .blend file but using the blender.app which is located in the same folder with the blend file, not with the blender.app which is located in Applications. (using Macosx)
So I was thinking that this should do the job...but instead it opens blender twice...
import os
path = os.getcwd()
print(path)
os.system("cd path/")
os.system("open blender.app Import_mhx.blend")
I also tried this one
import os
path = os.getcwd()
print(path)
os.system("cd path/")
os.system("open Import_mhx.blend")
but unfortunately it opens the .blend file with the default blender.app which is located in Applications...
any idea?
This cannot work since the system command gets executed in a subshell, and the chdir is only valid for that subshell. Replace the command by
os.system("open -a path/blender.app Import_mhx.blend")
or (much better)
subprocess.check_call(["open", "-a", os.path.join(path, "blender.app"),
"Import_mhx.blend"])
Have you tried telling the open command to open it WITH a specific application?
open -a /path/to/blender.app /path/to/Import_mhx.blend
Your first attempt was on the right track but you were really telling open to just open two different things. Not one with the other.