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'''
Copyright (C) 2007 Martin Owens
Thanks to Lineaire Chez of Inkbar ( www.inkbar.lineaire.net )
This program is free software; you can redistribute it and/or modify
it under the terms of the GNU General Public License as published by
the Free Software Foundation; either version 2 of the License, or
(at your option) any later version.
This program is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
GNU General Public License for more details.
You should have received a copy of the GNU General Public License
along with this program; if not, write to the Free Software
Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
'''
import EAN13
from EAN13 import mapLeftFaimly, guardBar, centerBar
import sys
mapFamily = [ '000111','001011','001101','001110','010011','011001','011100','010101','010110','011010' ]
class Object(EAN13.Object):
def encode(self, number):
result = ''
l = len(number)
if (l != 6 and l != 7 and l != 11 and l != 12) or not number.isdigit():
sys.stderr.write("Can not encode '" + number + "' into UPC-E Barcode, Size must be 6 numbers only, and 1 check digit (optional)\nOr a convertable 11 digit UPC-A number with 1 check digit (also optional).\n")
return
echeck = None
if l==7 or l==12:
echeck = number[-1]
number = number[:-1]
sys.stderr.write("CHECKSUM FOUND!")
l -= 1
if l==6:
number = self.ConvertEtoA(number)
if not echeck:
echeck = self.getChecksum(number)
else:
if not self.varifyChecksum(number + echeck):
sys.stderr.write("UPC-E Checksum not correct for this barcode, omit last charicter to generate new checksum.\n")
return
number = self.ConvertAtoE(number)
if not number:
sys.stderr.write("UPC-A code could not be converted into a UPC-E barcode, please follow the UPC guide or enter a 6 digit UPC-E number..\n")
return
number = number
result = result + guardBar
# The check digit isn't stored as bars but as a mirroring system. :-(
family = mapFamily[int(echeck)]
i = 0
for i in range(0,6):
result += mapLeftFaimly[int(family[i])-1][int(number[i])]
result = result + centerBar + '2';
self.label = '0 ' + number[:6] + ' ' + echeck
self.inclabel = self.label
return result;
def fontSize(self):
return '10'
def ConvertAtoE(self, number):
# Converting UPC-A to UPC-E
# All UPC-E Numbers use number system 0
if number[0] != '0' or len(number)!=11:
# If not then the code is invalid
return None
# Most of the conversions deal
# with the specific code parts
manufacturer = number[1:6]
product = number[6:11]
# There are 4 cases to convert:
if manufacturer[2:] == '000' or manufacturer[2:] == '100' or manufacturer[2:] == '200':
# Maxium number product code digits can be encoded
if product[:2]=='00':
return manufacturer[:2] + product[2:] + manufacturer[2]
elif manufacturer[3:5] == '00':
# Now only 2 product code digits can be used
if product[:3]=='000':
return manufacturer[:3] + product[3:] + '3'
elif manufacturer[4] == '0':
# With even more manufacturer code we have less room for product code
if product[:4]=='0000':
return manufacturer[0:4] + product[4] + '4'
elif product[:4]=='0000' and int(product[4]) > 4:
# The last recorse is to try and squeeze it in the last 5 numbers
# so long as the product is 00005-00009 so as not to conflict with
# the 0-4 used above.
return manufacturer + product[4]
else:
# Invalid UPC-A Numbe
return None
def ConvertEtoA(self, number):
# Convert UPC-E to UPC-A
# It's more likly to convert this without fault
# But we still must be mindful of the 4 conversions
if len(number)!=6:
return None
if number[5]=='0' or number[5]=='1' or number[5]=='2':
return '0' + number[:2] + number[5] + '0000' + number[2:5]
elif number[5]=='3':
return '0' + number[:3] + '00000' + number[3:5]
elif number[5]=='4':
return '0' + number[:4] + '00000' + number[4]
else:
return '0' + number[:5] + '0000' + number[5]
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