CkPython

Unzip Files Matching a Pattern (such as *.xml)

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Example code to unzip Files Matching a Pattern (such as *.xml) CkPython

import sys
import chilkat

success = False

# This example works with the .zip located at http://www.example-code.com/data/test.zip

# This example assumes the Chilkat API to have been previously unlocked.
# See Global Unlock Sample for sample code.

# First, let's download the .zip
http = chilkat.CkHttp()
success = http.Download("http://www.example-code.com/data/test.zip","qa_test/test.zip")
if (success != True):
    print(http.lastErrorText())
    sys.exit()

zip = chilkat.CkZip()

success = zip.OpenZip("qa_test/test.zip")
if (success != True):
    print(zip.lastErrorText())
    sys.exit()

# Show what's in the .zip by getting the directory as XML.
zipContentsXml = zip.getDirectoryAsXML()
if (zip.get_LastMethodSuccess() != True):
    print(zip.lastErrorText())
    sys.exit()

print(zipContentsXml)

# In this case, the contents of our .zip look like this:
# <?xml version="1.0" encoding="utf-8" ?>
# <zip_contents>
#     <dir name="test">
#         <dir name="abc">
#             <dir name="123">
#                 <file>ecc_public.pem</file>
#                 <file>starfish.jpg</file>
#             </dir>
#             <file>comodo-certificates.pem</file>
#             <file>pigs.xml</file>
#         </dir>
#         <file>COMODORSACertificationAuthority.crt</file>
#         <file>hamlet.xml</file>
#     </dir>
# </zip_contents>

# ------------------------------------------------------------------------
# If we wish to unzip only .xml files, use "*.xml" for the pattern\
# Unzip to the "qa_output" directory relative to our current working directory.
# (You may unzip to an absolute path by passing an absolute path, such as "/someDir" or
# "C:/someDir", etc.)
# This will unzip only the .xml files.  The relative directory structure within
# the .zip will be created in the local filesystem under the unzip directory
# path passed in the first argument, which in this case is "qa_output".
count = zip.UnzipMatching("qa_output","*.xml",True)
if (count < 0):
    print(zip.lastErrorText())
    sys.exit()

print("unzipped " + str(count) + " matching file(s).")

# ------------------------------------------------------------------------
# If we wish to unzip only files under the "123" directory, we can do this:
count = zip.UnzipMatching("qa_output2","*/123/*",True)
if (count < 0):
    print(zip.lastErrorText())
    sys.exit()

print("unzipped " + str(count) + " matching file(s).")

# ------------------------------------------------------------------------
# If no asterisk character is used, then the match must be exact.
# For example, to unzip "test/hamlet.xml" we must pass the exact path
# as it is in the .zip.  We cannot just pass "hamlet.xml".
count = zip.UnzipMatching("qa_output3","test/hamlet.xml",True)
if (count < 0):
    print(zip.lastErrorText())
    sys.exit()

print("unzipped " + str(count) + " matching file(s).")

# ------------------------------------------------------------------------
# If we want to unzip any file named "hamlet.xml" found in any sub-directory...
count = zip.UnzipMatching("qa_output4","*/hamlet.xml",True)
if (count < 0):
    print(zip.lastErrorText())
    sys.exit()

print("unzipped " + str(count) + " matching file(s).")