__init__.py question

[codecraig]

Ok, I have the following directory structure

C:\pycode
--> blah.py
--> mynewdir
--> __init__.py
--> abc.py

[[ C:\pycode\mynewdir\abc.py ]]

def doFoo():
print "hi"

def doBar():
print "bye"

[[ C:\pycode\mynewdir\__init__.py ]]

from mynewdir import *

[[ C:\pycode\blah.py ]]

????

what do i import in blah.py so that I can accesss, abc.doFoo() ?

thanks

[gry]

from mynewdir import abc
abc.doFoo()

or

import mynewdir.abc
newdir.abc.doFoo()

[Terry Hancock]

On Friday 22 April 2005 07:19 am, codecraig wrote:

Ok, I have the following directory structure

C:\pycode
--> blah.py
--> mynewdir
--> __init__.py
--> abc.py

[[ C:\pycode\mynewdir\abc.py ]]

def doFoo():
print "hi"

def doBar():
print "bye"

[[ C:\pycode\mynewdir\__init__.py ]]

from mynewdir import *
This didn't work, did it? There is no module
"mynewdir.py" nor a package "mynewdir" in
the "mynewdir" directory, and I don't think import
will search up to find the container.

I suspect you meant that __init__.py says:

from abc import *
[[ C:\pycode\blah.py ]]

????

what do i import in blah.py so that I can accesss,

abc.doFoo() ?

Assuming the above, and that you want to access
it as you have written it, that would be:

from mynewdir import abc

Note that in order to use this form, you don't have
to have *anything* in mynewdir/__init__.py --- it can
be an empty file, as long as it exists.

You only need to use an import in __init__.py if you
want it to automatically run when you import the
package.

E.g. if you did:

import mynewdir

You could access your function as:

mynewdir.abc.doFoo

(which requires the import statement in __init__.py).

Cheers,
Terry
--
Terry Hancock ( hancock at anansispaceworks.com )
Anansi Spaceworks http://www.anansispaceworks.com