Import problem
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From: News123 on 8 Mar 2010 17:21 Jean-Michel Pichavant wrote: > Johny wrote: >> I have this directory structure >> >> C: >> \A >> __init__.py >> amodule.py >> >> \B >> __init__.py >> bmodule.py >> >> \D >> __init__.py >> dmodule.py >> >> and I want to import bmodule.py >> C:\>cd \ >> >> C:\>python >> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >> (Intel)] on win >> 32 >> Type "help", "copyright", "credits" or "license" for more information. >> >>>>> from A.B import bmodule >>>>> >> I am bmodule >> C:\> >> >> so far so good. Now I would like to import bmodule but if the current >> directory is \D subdirectory. >> >> C:> cd \A\B\D >> C:\A\B\D> >> C:\A\B\D>python >> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >> (Intel)] on win >> 32 >> Type "help", "copyright", "credits" or "license" for more information. >> >>>>> import sys >>>>> sys.path.append('C:\\A') >>>>> from A.B import bmodule >>>>> >> Traceback (most recent call last): >> File "<stdin>", line 1, in <module> >> ImportError: No module named A.B >> >> C:\> >> >> so I can not import a module from the parent directory? Or where did I >> make an error? >> Thanks for help >> >> L. >> > try > > import sys > sys.path.append('C:\\') > from A.B import bmodule > is there any 'automatic' way of finding the top level directory?basically the 'top level directory is the first directory going upwards, that doesn't contain a __init__.py file. of course you could do this 'manually' by doing: # assume, that this module is A.amodule import sys import os # I'd love to have a similiar automatic construct if __name__ == "__main__": level = 1 # or function locating how far to go up before # finding a dir, whcih does not contain a __init__.py mydir = os.path.split(__file__)[0] topdir = os.path.join( mydir,*(("..",)*level)) abstop = os.path.abspath(topdir) sys.path.append(abstop) ## now you can import with the normal module paths import A.blo print "and I found blo",dir(A.blo) bye N
From: Jean-Michel Pichavant on 10 Mar 2010 07:27 News123 wrote: > Jean-Michel Pichavant wrote: > >> Johny wrote: >> >>> I have this directory structure >>> >>> C: >>> \A >>> __init__.py >>> amodule.py >>> >>> \B >>> __init__.py >>> bmodule.py >>> >>> \D >>> __init__.py >>> dmodule.py >>> >>> and I want to import bmodule.py >>> C:\>cd \ >>> >>> C:\>python >>> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >>> (Intel)] on win >>> 32 >>> Type "help", "copyright", "credits" or "license" for more information. >>> >>> >>>>>> from A.B import bmodule >>>>>> >>>>>> >>> I am bmodule >>> C:\> >>> >>> so far so good. Now I would like to import bmodule but if the current >>> directory is \D subdirectory. >>> >>> C:> cd \A\B\D >>> C:\A\B\D> >>> C:\A\B\D>python >>> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >>> (Intel)] on win >>> 32 >>> Type "help", "copyright", "credits" or "license" for more information. >>> >>> >>>>>> import sys >>>>>> sys.path.append('C:\\A') >>>>>> from A.B import bmodule >>>>>> >>>>>> >>> Traceback (most recent call last): >>> File "<stdin>", line 1, in <module> >>> ImportError: No module named A.B >>> >>> C:\> >>> >>> so I can not import a module from the parent directory? Or where did I >>> make an error? >>> Thanks for help >>> >>> L. >>> >>> >> try >> >> import sys >> sys.path.append('C:\\') >> from A.B import bmodule >> >> > is there any 'automatic' way of finding the top level > directory?basically the 'top level directory is the first directory > going upwards, that doesn't contain a __init__.py file. > what if some user has an __init__.py file the top level directory of your package ? > of course you could do this 'manually' by > doing: > > # assume, that this module is A.amodule > import sys > import os > > # I'd love to have a similiar automatic construct > if __name__ == "__main__": > level = 1 # or function locating how far to go up before > # finding a dir, whcih does not contain a __init__.py > mydir = os.path.split(__file__)[0] > topdir = os.path.join( mydir,*(("..",)*level)) > abstop = os.path.abspath(topdir) > sys.path.append(abstop) > > ## now you can import with the normal module paths > > import A.blo > print "and I found blo",dir(A.blo) > > > bye N > > > > You don't want to do that and you don't need it neither. That's what the env variable PYTHONPATH is for. set it correctly, install your package inside and everything works just fine (+standard). With a linux OS it easy to create smb links to point to any working directory. It should be possible on windows as well. If your package is meant to be destributed, you may use setup.py JM
From: News123 on 10 Mar 2010 18:17 Hi JM, Jean-Michel Pichavant wrote: > News123 wrote: >> Jean-Michel Pichavant wrote: >> >>> Johny wrote: >>> >>>> I have this directory structure >>>> >>>> C: >>>> \A >>>> __init__.py >>>> amodule.py >>>> >>>> \B >>>> __init__.py >>>> bmodule.py >>>> >>>> \D >>>> __init__.py >>>> dmodule.py >>>> >>>> and I want to import bmodule.py >>>> C:\>cd \ >>>> >>>> C:\>python >>>> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >>>> (Intel)] on win >>>> 32 >>>> Type "help", "copyright", "credits" or "license" for more information. >>>> >>>> >>>>>>> from A.B import bmodule >>>>>>> >>>> I am bmodule >>>> C:\> >>>> >>>> so far so good. Now I would like to import bmodule but if the current >>>> directory is \D subdirectory. >>>> >>>> C:> cd \A\B\D >>>> C:\A\B\D> >>>> C:\A\B\D>python >>>> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >>>> (Intel)] on win >>>> 32 >>>> Type "help", "copyright", "credits" or "license" for more information. >>>> >>>> >>>>>>> import sys >>>>>>> sys.path.append('C:\\A') >>>>>>> from A.B import bmodule >>>>>>> >>>> Traceback (most recent call last): >>>> File "<stdin>", line 1, in <module> >>>> ImportError: No module named A.B >>>> >>>> C:\> >>>> >>>> so I can not import a module from the parent directory? Or where did I >>>> make an error? >>>> Thanks for help >>>> >>>> L. >>>> >>> try >>> >>> import sys >>> sys.path.append('C:\\') >>> from A.B import bmodule >>> >>> >> is there any 'automatic' way of finding the top level >> directory?basically the 'top level directory is the first directory >> going upwards, that doesn't contain a __init__.py file. >> > what if some user has an __init__.py file the top level directory of > your package ? Is there any other usage of __init.py__ than indicating a module directory? I wasn't aware of it, but you're right I did not investigte in depth and users can of course do whatever they like. >> of course you could do this 'manually' by >> doing: >> >> # assume, that this module is A.amodule >> import sys >> import os >> >> # I'd love to have a similiar automatic construct >> if __name__ == "__main__": >> level = 1 # or function locating how far to go up before >> # finding a dir, whcih does not contain a __init__.py >> mydir = os.path.split(__file__)[0] >> topdir = os.path.join( mydir,*(("..",)*level)) >> abstop = os.path.abspath(topdir) >> sys.path.append(abstop) >> >> ## now you can import with the normal module paths >> >> import A.blo >> print "and I found blo",dir(A.blo) >> >> > You don't want to do that and you don't need it neither. That's what the > env variable PYTHONPATH is for. set it correctly, install your package > inside and everything works just fine (+standard). With a linux OS it > easy to create smb links to point to any working directory. It should be > possible on windows as well. I like your idea with the symlinks. However not sure how to do it with windows. I assume default shortcuts won't do. > > If your package is meant to be destributed, you may use setup.py > > Well, It's nice if a user just unpacks a zip file and can click on any script with the .py suffix in the tree. (Its nice for example for tutorials / demos ) It's also nice if he can later on just delete the unpacked directory and there will be no trace left in the registry or in the python base dir. This is why I'm interested in solutions without setup.py or changing environment variables. bye N
From: News123 on 10 Mar 2010 18:18
Hi JM, Jean-Michel Pichavant wrote: > News123 wrote: >> Jean-Michel Pichavant wrote: >> >>> Johny wrote: >>> >>>> I have this directory structure >>>> >>>> C: >>>> \A >>>> __init__.py >>>> amodule.py >>>> >>>> \B >>>> __init__.py >>>> bmodule.py >>>> >>>> \D >>>> __init__.py >>>> dmodule.py >>>> >>>> and I want to import bmodule.py >>>> C:\>cd \ >>>> >>>> C:\>python >>>> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >>>> (Intel)] on win >>>> 32 >>>> Type "help", "copyright", "credits" or "license" for more information. >>>> >>>> >>>>>>> from A.B import bmodule >>>>>>> >>>> I am bmodule >>>> C:\> >>>> >>>> so far so good. Now I would like to import bmodule but if the current >>>> directory is \D subdirectory. >>>> >>>> C:> cd \A\B\D >>>> C:\A\B\D> >>>> C:\A\B\D>python >>>> Python 2.5 (r25:51908, Sep 19 2006, 09:52:17) [MSC v.1310 32 bit >>>> (Intel)] on win >>>> 32 >>>> Type "help", "copyright", "credits" or "license" for more information. >>>> >>>> >>>>>>> import sys >>>>>>> sys.path.append('C:\\A') >>>>>>> from A.B import bmodule >>>>>>> >>>> Traceback (most recent call last): >>>> File "<stdin>", line 1, in <module> >>>> ImportError: No module named A.B >>>> >>>> C:\> >>>> >>>> so I can not import a module from the parent directory? Or where did I >>>> make an error? >>>> Thanks for help >>>> >>>> L. >>>> >>> try >>> >>> import sys >>> sys.path.append('C:\\') >>> from A.B import bmodule >>> >>> >> is there any 'automatic' way of finding the top level >> directory?basically the 'top level directory is the first directory >> going upwards, that doesn't contain a __init__.py file. >> > what if some user has an __init__.py file the top level directory of > your package ? Is there any other usage of __init.py__ than indicating a module directory? I wasn't aware of it, but you're right I did not investigte in depth and users can of course do whatever they like. >> of course you could do this 'manually' by >> doing: >> >> # assume, that this module is A.amodule >> import sys >> import os >> >> # I'd love to have a similiar automatic construct >> if __name__ == "__main__": >> level = 1 # or function locating how far to go up before >> # finding a dir, whcih does not contain a __init__.py >> mydir = os.path.split(__file__)[0] >> topdir = os.path.join( mydir,*(("..",)*level)) >> abstop = os.path.abspath(topdir) >> sys.path.append(abstop) >> >> ## now you can import with the normal module paths >> >> import A.blo >> print "and I found blo",dir(A.blo) >> >> > You don't want to do that and you don't need it neither. That's what the > env variable PYTHONPATH is for. set it correctly, install your package > inside and everything works just fine (+standard). With a linux OS it > easy to create smb links to point to any working directory. It should be > possible on windows as well. I like your idea with the symlinks. However not sure how to do it with windows. I assume default shortcuts won't do. > > If your package is meant to be destributed, you may use setup.py > > Well, It's nice if a user just unpacks a zip file and can click on any script with the .py suffix in the tree. (Its nice for example for tutorials / demos ) It's also nice if he can later on just delete the unpacked directory and there will be no trace left in the registry or in the python base dir. This is why I'm interested in solutions without setup.py or changing environment variables. bye N |