Author Topic: RC3 Python debugging  (Read 1186 times)

jporkkahtc

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RC3 Python debugging
« on: September 23, 2016, 10:17:40 pm »
Seems to be busted (or at least, unintuitive)
I created a new empty python project (in a new workspace), added my single *.py file to it.
Then, step-in. In "Build", I get this message:
C:\bin>Traceback (most recent call last):
  File "C:\Program Files\SlickEdit Pro 21.0.0\resource\tools\pydbgp-1.2.0\pythonlib\dbgp\client.py", line 2441, in runMain
    self.dbg.runfile(debug_args[0], debug_args)
  File "C:\Program Files\SlickEdit Pro 21.0.0\resource\tools\pydbgp-1.2.0\pythonlib\dbgp\client.py", line 2028, in runfile
    h_execfile(file, args, module=main, tracer=self)
  File "C:\Program Files\SlickEdit Pro 21.0.0\resource\tools\pydbgp-1.2.0\pythonlib\dbgp\client.py", line 689, in __init__
    execfile(file, globals, locals)
IOError: [Errno 21] Is a directory: 'C:\\bin'


The problem, I hadn't actually *opened* the *.py file -- so I had the new project but no files open.
In same configuration, press F5 to start debugging and you get:
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SlickEdit Pro
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".command" does not appear to be a runnable script.

Continue to execute this script?

Note: You can choose a default script to run from the Options dialog ("Build", "Python Options")
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&Yes   &No   
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HM... Pressing F5 behaves differently than picking the Debug->Start menu.
The menu actuallys starts the debugger - but there isn't anything to debug.

The python process is started with "" as the script to debug:
python  -S -u "C:\Program Files\SlickEdit Pro 21.0.0\resource\tools\pydbgp-1.2.0/bin/pydbgp_bootstrapper.py" -d 127.0.0.1:55622 -k slickedit ""