I would like to get just the folder path from the full path to a file.
For example T:\Data\DBDesign\DBDesign_93_v141b.mdb
and I would like to get just T:\Data\DBDesign
(excluding the \DBDesign_93_v141b.mdb
).
I have tried something like this:
existGDBPath = r'T:\Data\DBDesign\DBDesign_93_v141b.mdb'
wkspFldr = str(existGDBPath.split('\\')[0:-1])
print wkspFldr
but it gave me a result like this:
['T:', 'Data', 'DBDesign']
which is not the result that I require (being T:\Data\DBDesign
).
Any ideas on how I can get the path to my file?
You were almost there with your use of the split
function. You just needed to join the strings, like follows.
>>> import os
>>> '\\'.join(existGDBPath.split('\\')[0:-1])
'T:\\Data\\DBDesign'
Although, I would recommend using the os.path.dirname
function to do this, you just need to pass the string, and it'll do the work for you. Since, you seem to be on windows, consider using the abspath
function too. An example:
>>> import os
>>> os.path.dirname(os.path.abspath(existGDBPath))
'T:\\Data\\DBDesign'
If you want both the file name and the directory path after being split, you can use the os.path.split
function which returns a tuple, as follows.
>>> import os
>>> os.path.split(os.path.abspath(existGDBPath))
('T:\\Data\\DBDesign', 'DBDesign_93_v141b.mdb')