用于Code128条形码字体的Python Code128编码器

Question

就像标题上写的，我有一个Code128字体，我想用它打印条形码。但是，字符串需要用Code128编码才能使条形码字体工作。我的应用程序使用Python3语言。

曾经有一个例子在网络上某处如何编码字符串为Code128字体，但我找不到它了。

我做而不是想要一个字符串到.svg转换器。我特别希望将字符串转换为Code128编码的字符串。

Python3中的任何引用、代码片段或文档都将不胜感激。

编辑:我使用来自这里的字体。

Answer 1

这是一个公认的答案，所以我把原来的代码留在下面。但我更喜欢这种精致。

def list_join(seq):
    ''' Join a sequence of lists into a single list, much like str.join
        will join a sequence of strings into a single string.
    '''
    return [x for sub in seq for x in sub]

code128B_mapping = dict((chr(c), [98, c+64] if c < 32 else [c-32]) for c in range(128))
code128C_mapping = dict([(u'%02d' % i, [i]) for i in range(100)] + [(u'%d' % i, [100, 16+i]) for i in range(10)])
code128_chars = u''.join(chr(c) for c in [212] + list(range(33,126+1)) + list(range(200,211+1)))

def encode128(s):
    ''' Code 128 conversion for a font as described at
        https://en.wikipedia.org/wiki/Code_128 and downloaded
        from http://www.barcodelink.net/barcode-font.php
        Only encodes ASCII characters, does not take advantage of
        FNC4 for bytes with the upper bit set. Control characters
        are not optimized and expand to 2 characters each.
        Coded for https://stackoverflow.com/q/52710760/5987
    '''
    if s.isdigit() and len(s) >= 2:
        # use Code 128C, pairs of digits
        codes = [105] + list_join(code128C_mapping[s[i:i+2]] for i in range(0, len(s), 2))
    else:
        # use Code 128B and shift for Code 128A
        codes = [104] + list_join(code128B_mapping[c] for c in s)
    check_digit = (codes[0] + sum(i * x for i,x in enumerate(codes))) % 103
    codes.append(check_digit)
    codes.append(106) # stop code
    return u''.join(code128_chars[x] for x in codes)

def encode128(s):
    ''' Code 128 conversion for a font as described at
        https://en.wikipedia.org/wiki/Code_128 and downloaded
        from http://www.barcodelink.net/barcode-font.php
        Only encodes ASCII characters, does not take advantage of
        FNC4 for bytes with the upper bit set.
        It does not attempt to optimize the length of the string,
        Code B is the default to prefer lower case over control characters.
        Coded for https://stackoverflow.com/q/52710760/5987
    '''
    s = s.encode('ascii').decode('ascii')
    if s.isdigit() and len(s) % 2 == 0:
        # use Code 128C, pairs of digits
        codes = [105]
        for i in range(0, len(s), 2):
            codes.append(int(s[i:i+2], 10))
    else:
        # use Code 128B and shift for Code 128A
        mapping = dict((chr(c), [98, c + 64] if c < 32 else [c - 32]) for c in range(128))
        codes = [104]
        for c in s:
            codes.extend(mapping[c])
    check_digit = (codes[0] + sum(i * x for i,x in enumerate(codes))) % 103
    codes.append(check_digit)
    codes.append(106) # stop code
    chars = (b'\xd4' + bytes(range(33,126+1)) + bytes(range(200,211+1))).decode('latin-1')
    return ''.join(chars[x] for x in codes)