vieyahn2017
commented
3 years ago 首先，可以通过python代码解密读取出vbaProject.bin
#!/usr/bin/python
##############################################################################
# vba_extract.py
#
# vba_extract - A simple utility to extract a vbaProject.bin binary from an
# Excel 2007+ xlsm file for insertion into an XlsxWriter file.
#
# Copyright 2013-2016, John McNamara, jmcnamara@cpan.org
#
import sys
import shutil
from zipfile import ZipFile
from zipfile import BadZipfile
# The VBA project file we want to extract.
vba_filename = 'vbaProject.bin'
#import pdb
#pdb.set_trace()
# Get the xlsm file name from the commandline.
if len(sys.argv) > 1:
xlsm_file = sys.argv[1]
else:
xlsm_file = "11.xlsm"
try:
# Open the Excel xlsm file as a zip file.
xlsm_zip = ZipFile(xlsm_file, 'r')
# print(xlsm_zip.namelist())
# xlsm_zip.extractall()
# xlsm_zip.extract('xl/vbaProject.bin')
# Read the xl/vbaProject.bin file.
vba_data = xlsm_zip.read('xl/' + vba_filename)
# Write the vba data to a local file.
vba_file = open(vba_filename, "wb")
vba_file.write(vba_data)
vba_file.close()
except IOError:
# Use exc_info() for Python 2.5+ compatibility.
e = sys.exc_info()[1]
print("File error: %s" % str(e))
exit()
except KeyError:
# Usually when there isn't a xl/vbaProject.bin member in the file.
e = sys.exc_info()[1]
print("File error: %s" % str(e))
print("File may not be an Excel xlsm macro file: '%s'" % xlsm_file)
exit()
except BadZipfile:
# Usually if the file is an xls file and not an xlsm file.
e = sys.exc_info()[1]
print("File error: %s: '%s'" % (str(e), xlsm_file))
print("File may not be an Excel xlsm macro file.")
exit()
except:
# Catch any other exceptions.
e = sys.exc_info()[1]
print("File error: %s" % str(e))
exit()
finally:
xlsm_zip.close()
print("Extracted: %s" % vba_filename)
Excel的vba解密
Excel文件可以以zip方式打开，但是其中的vba代码是以加密方式存储的， 具体代码加密存储在 xl/vbaProject.bin