Repository navigation
Cannot get target from remote resource in python shell #1566
Unanswered
nvincent-vossloh
asked this question in
Q&A
Replies: 3 comments 3 replies
|
Which version of labgrid are you using? |
1 reply
|
Hello, I am using custom version available at https://github.com/SiemaApplications-attic/labgrid/tree/dda69a2515d18396816a97f63ebcc63bcae8afe4 $ labgrid-client version
23.1a3.dev195pip freeze output: |
2 replies
|
Unfortunately it does not work either with the vanilla version of labgrid ( I am wondering if the version of the coordinator and/or the |
0 replies
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Uh oh!
There was an error while loading. Please reload this page.
Hello,
After being away from labgrid I came back recently to integrate some labgrid features into our test environment.
I used to debug with a python shell open and test various commands with the following setup:
(from a python shell)
Nothing fancy, it comes from the online documentation https://labgrid.readthedocs.io/en/stable/usage.html#environments
the environment file looks like this
(
StLinkV3Driveris a custom labgrid driver I implemented when I did not how to useopenocd, it works fine for my use case and is already integrated in our pytest environment.)The issue I am having is when I hit
t = e.get_target('main')I have the following error:I have defined and exported an
LG_CROSSBARenvironement var.I have defined and exported
no_proxy/NO_PROXYvar to set the ip of the coordinator and the exporter.When I provide a 'local' environment file (one where resources are in the file instead of a
RemotePlaceresource) the command succeeds. But as you can imagine, I need to test with the coordinator/exporter in the loop.Looking at my ipython history, it looks like I had managed to get the
get_targetto return something that I could use.It is probably something really silly that I forgot, but I cannot see what 🙁
Any help would be greatly appreciated 🙂
All reactions