Python Code128条码字体的Code128编码器

Question

正如标题所说，我有一个 Code128 字体，我想用它来打印条形码。但是，字符串需要以 Code128 编码才能使条形码字体正常工作。我的应用程序使用 Python3 语言。

网络上曾经有一个示例如何将字符串编码为 Code128 字体，但我找不到了。

我不想要一个字符串到 .svg 的转换器。我特别想将字符串转换为Code128编码的字符串。

Python3 或文档中的任何参考、代码片段将不胜感激。

编辑：我使用 here 中的字体。

Answer 1

这是一个公认的答案，所以我将原始代码留在下面。但我更喜欢这种细化。

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 
    '''
    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 
    '''
    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)

Python Code128条码字体的Code128编码器

Python Code128 encoder for Code128 barcode font

python

barcode

code128

python-3.x