Python interface how to detect file is not open?

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7 years 8 months ago #126037 by bladekel
Hello to all...

I'm trying to create a custom ui with python and I need to detect if an ngc file couldnt opened succesfully.

For this problem I'm using interpreter_errcode like;

self.s = self.emc.stat()

self.s.interpreter_errcode

but there is no return with this code..

Can anyone help me?

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7 years 8 months ago #126043 by BigJohnT
Code:
#!/usr/bin/env python # -*- coding: utf-8 -*- import sys import linuxcnc try: s = linuxcnc.stat() # create a connection to the status channel s.poll() # get current values except linuxcnc.error, detail: print "error", detail sys.exit(1) s.file # returns the current loaded file

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7 years 8 months ago #126052 by cmorley
I might add that you should check the file name (using linuxcnc status) before you load the program, then check it after. If they are the same then the program probably loaded file.

eg gladevcp doe it like this (the code wont work for you but the idea is good):

def _load_file(self, filename):
if filename:
self.linuxcnc.mode(linuxcnc.MODE_AUTO)
old = self.gstat.stat.file
self.linuxcnc.program_open(filename)
if old == filename:
self.gstat.emit('file-loaded', filename)

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7 years 8 months ago #126192 by bladekel
Thanks bro. Thats the way....

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