CODEWITHSUNDEEP
pythonpysparkazure-databricks
Anonymous
Anonymous

Reputation: 329

How to specify wildcard filenames for .Zip type files in Python Variables

My requirement is

I'm calling a function to extract files/folders from a .Zip file

files_extract_with_structure(file_source, file_dest)

file_source and file_dest are the variables I'm passing to the above function and the value of the file_source variable is defined as below.

file_source = "/dbfs/mnt/devadls/pre/Source_Files/2022-10/767676.XXX.XXX.XXXX.20221010090858.txt.zip"

where 767676.XXX.XXX.XXXX.20221010090858.txt.zip is the zip file name

The above function works fine if I pass the file_source variable value as above (hardcoded the zip file name)

My requirement is instead of hard coding the zip file name, Can we specify the wild card file names as below?

file_source = "/dbfs/mnt/devadls/pre/Source_Files/2022-10/767676.XXX.XXX.XXXX.*.txt.zip"

because I will receive the same file with a different date in the next month and so on...

But when I specify the wildcard names as "767676.XXX.XXX.XXXX.*.txt.zip", it is throwing a `No such file or directory error.

Kindly help to resolve this issue. Thanks.

Upvotes: 0

Views: 251

Answers (1)

Tamles
Tamles

Reputation: 66

You could use the fnmatch module from the standard library, it will let you filter and match filename based on unix-like rules.

With your example, something like this should work:

from fnmatch import fnmatch

file_source = "/dbfs/mnt/devadls/pre/Source_Files/2022-10/767676.XXX.XXX.XXXX.20221010090858.txt.zip"
pattern = "/dbfs/mnt/devadls/pre/Source_Files/2022-10/767676.XXX.XXX.XXXX.*.txt.zip"


if fnmatch(file_source, pattern):
    files_extract_with_structure(file_source, file_dest)
else:
    print("No file found")

The next step might be to list the files in the source directory, and for each to test against the pattern.

Hope this will help!

Upvotes: 2

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