Getting folder of current bot...

[Stuart Axon]

A lot of examples have hardcoded paths in them, I'm looking to avoid this...

I've got __file__ working in my branch, so you can do

os.path.dirname(__file__) + 'someresource.svg'


Any objections to me adding
__dirname__ with the current folder ?

This would make the code simpler + it is consistant with __file__ being a builtin.

S++

[Stuart Axon]

Having said that, maybe we should have an api equivilent to getResource in java maybe?

S++



----- Original Message ----

Sent: Tue, June 1, 2010 9:22:06 PM
Subject: Getting folder of current bot...
A lot of examples have hardcoded paths in them, I'm looking to avoid
this...

I've got __file__ working in my branch, so you can

do

os.path.dirname(__file__) + 'someresource.svg'


Any

objections to me adding

__dirname__ with the current folder

?

This would make the code simpler + it is consistant with __file__ being

a builtin.

S++

[ricardo lafuente]

Sounds fine -- we'll just have to document it along with the other
additions to Nodebox.

Post by Stuart Axon
A lot of examples have hardcoded paths in them, I'm looking to avoid this...
I've got __file__ working in my branch, so you can do
os.path.dirname(__file__) + 'someresource.svg'
Any objections to me adding
__dirname__ with the current folder ?
This would make the code simpler + it is consistant with __file__ being a builtin.
S++
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[Dave Crossland]

Post by Stuart Axon
This would make the code simpler + it is consistant with __file__ being a builtin.

+1