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Copyright (C) 2007 Martin Owens
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Thanks to Lineaire Chez of Inkbar ( www.inkbar.lineaire.net )
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This program is free software; you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation; either version 2 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program; if not, write to the Free Software
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Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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from EAN13 import mapLeftFaimly, guardBar, centerBar
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# Copyright (C) 2010 Martin Owens
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# This program is free software; you can redistribute it and/or modify
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# it under the terms of the GNU General Public License as published by
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# the Free Software Foundation; either version 2 of the License, or
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# (at your option) any later version.
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# This program is distributed in the hope that it will be useful,
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# but WITHOUT ANY WARRANTY; without even the implied warranty of
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# MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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# GNU General Public License for more details.
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# You should have received a copy of the GNU General Public License
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# along with this program; if not, write to the Free Software
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# Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA
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Python barcode renderer for UPCE barcodes. Designed for use with Inkscape.
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from BaseEan import EanBarcode
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mapFamily = [ '000111','001011','001101','001110','010011','011001','011100','010101','010110','011010' ]
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class Object(EAN13.Object):
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def encode(self, number):
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if (l != 6 and l != 7 and l != 11 and l != 12) or not number.isdigit():
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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")
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sys.stderr.write("CHECKSUM FOUND!")
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number = self.ConvertEtoA(number)
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echeck = self.getChecksum(number)
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if not self.verifyChecksum(number + echeck):
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sys.stderr.write("UPC-E Checksum not correct for this barcode, omit last character to generate new checksum.\n")
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number = self.ConvertAtoE(number)
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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")
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result = result + guardBar
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# The check digit isn't stored as bars but as a mirroring system. :-(
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family = mapFamily[int(echeck)]
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result += mapLeftFaimly[int(family[i])-1][int(number[i])]
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result = result + centerBar + '2';
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self.label = '0 ' + number[:6] + ' ' + echeck
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self.inclabel = self.label
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def ConvertAtoE(self, number):
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# Converting UPC-A to UPC-E
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# All UPC-E Numbers use number system 0
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if number[0] != '0' or len(number)!=11:
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# If not then the code is invalid
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# Most of the conversions deal
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# with the specific code parts
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manufacturer = number[1:6]
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product = number[6:11]
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# There are 4 cases to convert:
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if manufacturer[2:] == '000' or manufacturer[2:] == '100' or manufacturer[2:] == '200':
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# Maxium number product code digits can be encoded
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return manufacturer[:2] + product[2:] + manufacturer[2]
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elif manufacturer[3:5] == '00':
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# Now only 2 product code digits can be used
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if product[:3]=='000':
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return manufacturer[:3] + product[3:] + '3'
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elif manufacturer[4] == '0':
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# With even more manufacturer code we have less room for product code
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if product[:4]=='0000':
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return manufacturer[0:4] + product[4] + '4'
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elif product[:4]=='0000' and int(product[4]) > 4:
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# The last recorse is to try and squeeze it in the last 5 numbers
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# so long as the product is 00005-00009 so as not to conflict with
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# the 0-4 used above.
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return manufacturer + product[4]
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# Invalid UPC-A Numbe
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def ConvertEtoA(self, number):
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# Convert UPC-E to UPC-A
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# It's more likly to convert this without fault
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# But we still must be mindful of the 4 conversions
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if number[5]=='0' or number[5]=='1' or number[5]=='2':
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return '0' + number[:2] + number[5] + '0000' + number[2:5]
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return '0' + number[:3] + '00000' + number[3:5]
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return '0' + number[:4] + '00000' + number[4]
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return '0' + number[:5] + '0000' + number[5]
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# This is almost exactly the same as the standard FAMILIES
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# But flipped around and with the first 111000 instead of 000000.
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FAMS = [ '111000', '110100', '110010', '110001', '101100',
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'100110', '100011', '101010', '101001', '100101' ]
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class Object(EanBarcode):
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"""Generate EAN6/UPC-E barcode generator"""
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"""Generate a UPC-E Barcode"""
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self.label = self.space(['0'], 2, n[:6], 2, n[-1])
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code = self.encode_interleaved(n[-1], n[:6], FAMS)
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# 202(guard) + code + 020(center) + 202(guard)
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return self.enclose(code, center='020')
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def appendChecksum(self, number):
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"""Generate a UPCE Checksum"""
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number = self.ConvertEtoA(number)
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result = self.getChecksum(number)
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return self.ConvertAtoE(number) + result
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"""We need a font size of 10"""
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def ConvertAtoE(self, number):
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"""Converting UPC-A to UPC-E, may cause errors."""
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# All UPC-E Numbers use number system 0
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if number[0] != '0' or len(number)!=11:
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# If not then the code is invalid
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# Most of the conversions deal
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# with the specific code parts
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manufacturer = number[1:6]
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product = number[6:11]
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# There are 4 cases to convert:
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if manufacturer[2:] == '000' or manufacturer[2:] == '100' or manufacturer[2:] == '200':
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# Maxium number product code digits can be encoded
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return manufacturer[:2] + product[2:] + manufacturer[2]
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elif manufacturer[3:5] == '00':
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# Now only 2 product code digits can be used
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if product[:3]=='000':
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return manufacturer[:3] + product[3:] + '3'
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elif manufacturer[4] == '0':
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# With even more manufacturer code we have less room for product code
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if product[:4]=='0000':
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return manufacturer[0:4] + product[4] + '4'
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elif product[:4]=='0000' and int(product[4]) > 4:
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# The last recorse is to try and squeeze it in the last 5 numbers
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# so long as the product is 00005-00009 so as not to conflict with
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return manufacturer + product[4]
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def ConvertEtoA(self, number):
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"""Convert UPC-E to UPC-A by padding with zeros"""
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# It's more likly to convert this without fault
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# But we still must be mindful of the 4 conversions
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if number[5] in ['0', '1', '2']:
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return '0' + number[:2] + number[5] + '0000' + number[2:5]
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elif number[5] == '3':
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return '0' + number[:3] + '00000' + number[3:5]
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elif number[5] == '4':
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return '0' + number[:4] + '00000' + number[4]
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return '0' + number[:5] + '0000' + number[5]
added
removed
share/extensions/Barcode/UPCE.py
