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Python解析Word文档的自动编号_python 获取wps编号
作者：码创造者 | 2024-08-19 09:18:36
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python 获取wps编号
关于自动编号的知识可以参考《在 Open XML WordprocessingML 中使用编号列表》
链接：https://learn.microsoft.com/zh-cn/previous-versions/office/ee922775(v=office.14)
python-docx库并不能直接解析出Word文档的自动编号，因为原理较为复杂，但我们希望python能够读取自动编号对应的文本。
基本解析原理
为了测试验证，我们创建一个带有编号的文档进行测试，例如：
然后我们先看看主文档中，对应的xml存储：
from docx import Document
doc = Document(r"编号测试1.docx")
for paragraph in doc.paragraphs:
print(paragraph._element.xml)
break
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结果：
<w:p ...> <w:pPr> <w:numPr> <w:ilvl w:val="0"/> <w:numId w:val="1"/> </w:numPr> <w:bidi w:val="0"/> <w:ind w:left="0" w:leftChars="0" w:firstLine="0" w:firstLineChars="0"/> <w:rPr> <w:rFonts w:hint="eastAsia"/> <w:lang w:val="en-US" w:eastAsia="zh-CN"/> </w:rPr> </w:pPr> <w:r> <w:rPr> <w:rFonts w:hint="eastAsia"/> <w:lang w:val="en-US" w:eastAsia="zh-CN"/> </w:rPr> <w:t>第一章</w:t> </w:r> </w:p>
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在微软的文档中，说明了最重要的部分：
w:numPr 元素包含自动编号元素。w:ilvl 元素从零开始表示编号等级，w:numId 元素是编号部件的索引。
w:numId 为 0 值时 ，表示编号已经被删除段落不含列表项。
所以我们可以根据段落是否存在w:numPr并且w:numId的值不为0判断段落是否存在自动编号。
然后我们需要获取每个w:numId对应的自动编号状态，这个信息存储在zip压缩包的\word\numbering.xml文件中，可以参考微软文档的示例：
w:numbering同时包含w:num和w:abstractNum两种节点，其中w:num记录了 每个numId对应的abstractNumId，而w:abstractNum记录了每个abstractNumId对应的编号格式，包含了每个级别的编号样式信息。对于w:num，python-docx库已经帮我们解析好，可以直接读取，但w:abstractNum节点python-docx库却并未进行解析，只能我们自己进行xml解析。
可以通过如下代码获取每个numId对应的abstractNumId：
from docx import Document
doc = Document(r"编号测试1.docx")
numbering_part = doc.part.numbering_part._element
numId2abstractId = {
num.numId: num.abstractNumId.val for num in numbering_part.num_lst
}
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接下来我们需要解析w:abstractNum节点，查阅python-docx库的源码可以知道，它使用lxml的etree进行xml解析。
初步解析代码为：
from docx.oxml.ns import qn
abstractNumId2style = {}
for abstractNumIdTag in numbering_part.findall(qn("w:abstractNum")):
abstractNumId = abstractNumIdTag.get(qn("w:abstractNumId"))
for lvlTag in abstractNumIdTag.findall(qn("w:lvl")):
ilvl = lvlTag.get(qn("w:ilvl"))
style = {tag.tag[tag.tag.rfind("}") + 1:]: tag.get(qn("w:val"))
for tag in lvlTag.xpath("./*[@w:val]", namespaces=numbering_part.nsmap)}
abstractNumId2style[(int(abstractNumId), int(ilvl))] = style
print(abstractNumId2style)
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注意：docx.oxml.ns的qn函数可以将w:转换为对应的命名空间名称，但对于xpath表达式却无法正确处理，所以对于xpath表达式使用namespaces传入对应的命名空间。
除了上面的解析方法以外，还可以事先将节点的所有命名空间清除后再解析，清除代码如下：
def remove_namespace(node): node_tag = node.tag if '}' in node_tag: node.tag = node_tag[node_tag.rfind("}") + 1:] for attr_key in list(node.attrib): if '}' in attr_key: new_attr_key = attr_key[attr_key.rfind("}") + 1:] node.attrib[new_attr_key] = node.attrib.pop(attr_key) for child in node: remove_namespace(child) return node'
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运行这样可以递归消除目标节点所有子节点的命名空间。
可以每个类别每个级别的自动编号的属性信息：
{(0, 0): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.', 'lvlJc': 'left'}, (0, 1): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.', 'lvlJc': 'left'}, (0, 2): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.', 'lvlJc': 'left'}, (0, 3): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.%4.', 'lvlJc': 'left'}, (0, 4): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.%4.%5.', 'lvlJc': 'left'}, (0, 5): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.%4.%5.%6.', 'lvlJc': 'left'}, (0, 6): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.%4.%5.%6.%7.', 'lvlJc': 'left'}, (0, 7): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.%4.%5.%6.%7.%8.', 'lvlJc': 'left'}, (0, 8): {'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.%4.%5.%6.%7.%8.%9.', 'lvlJc': 'left'}}
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当然我们只测试了最基本的数值型自动编号，有些自动编号对应的节点没有直接的w:numFmt节点，解析代码还需针对性调整。
微软的文档中提到，对多级列表的某一级列表进行特殊设定时，w:num内会出现w:lvlOverride节点，但本人使用wps反复测试过后并没有出现。估计这种格式的xml只会在老版的office中出现，而且我们也不会故意在多级列表的某一级进行特殊设定，所以我们不考虑这种情况。
还需要考虑 w:suff 元素控制的列表后缀，即列表项与段落之间的空白内容，有可能为制表符和空格，也可以什么都没有。处理代码为：
{"space": " ", "nothing": ""}.get(style.get("suff"), "\t")
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多级编号处理
首先尝试读取每个段落对应的自动编号样式：
for paragraph in doc.paragraphs:
numpr = paragraph._element.pPr.numPr
if numpr is not None and numpr.numId.val != 0:
numId = numpr.numId.val
ilvl = numpr.ilvl.val
abstractId = numId2abstractId[numId]
style = abstractNumId2style[(abstractId, ilvl)]
print(style)
print(paragraph.text)
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结果：
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.', 'lvlJc': 'left'}
第一章
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.', 'lvlJc': 'left'}
第一节
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.', 'lvlJc': 'left'}
第二节
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.', 'lvlJc': 'left'}
第一条
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.%2.%3.', 'lvlJc': 'left'}
第二条
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.', 'lvlJc': 'left'}
第二章
{'start': '1', 'numFmt': 'decimal', 'lvlText': '%1.', 'lvlJc': 'left'}
第三章
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我们需要一个计数器来记录每个样式出现的次数，从而生成其对应的编号。
cache = {} for paragraph in doc.paragraphs: numpr = paragraph._element.pPr.numPr lvlText = "" if numpr is not None and numpr.numId.val != 0: numId = numpr.numId.val ilvl = numpr.ilvl.val abstractId = numId2abstractId[numId] style = abstractNumId2style[(abstractId, ilvl)] if (abstractId, ilvl) in cache: cache[(abstractId, ilvl)] += 1 else: cache[(abstractId, ilvl)] = int(style["start"]) lvlText = style.get("lvlText") for i in range(0, ilvl + 1): lvlText = lvlText.replace(f'%{i + 1}', str(cache[(abstractId, i)])) suff_text = {"space": " ", "nothing": ""}.get(style.get("suff"), "\t") lvlText += suff_text print(lvlText + paragraph.text)
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结果：
1. 第一章
1.1. 第一节
1.2. 第二节
1.2.1. 第一条
1.2.2. 第二条
2. 第二章
3. 第三章
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各种其他类型的编号生成
为了尽量多的支持更多类型的编号，我创建了如下测试文件：
我们没有必要获取对应的圆圈数字，圆圈就获取对应的整数。
除了三种日文编号，上面的示例几乎包含所有的编号类型。需要注意三位数以上的数字格式，其xml有些特殊，例如：
<w:lvl> <w:start w:val="1"/> <mc:AlternateContent> <mc:Choice Requires="w14"> <w:numFmt w:val="custom" w:format="001, 002, 003, ..."/> </mc:Choice> <mc:Fallback> <w:numFmt w:val="decimal"/> </mc:Fallback> </mc:AlternateContent> <w:suff w:val="space"/> <w:lvlText w:val="%1"/> <w:lvlJc w:val="left"/> <w:pPr> <w:tabs> <w:tab w:val="left" w:pos="0"/> </w:tabs> </w:pPr> <w:rPr> <w:rFonts w:hint="default"/> </w:rPr> </w:lvl>
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基于此，解析格式的代码也作出如下调整：
abstractNumId2style = {} for abstractNumIdTag in numbering_part.findall(qn("w:abstractNum")): abstractNumId = abstractNumIdTag.get(qn("w:abstractNumId")) for lvlTag in abstractNumIdTag.findall(qn("w:lvl")): ilvl = lvlTag.get(qn("w:ilvl")) style = {tag.tag[tag.tag.rfind("}") + 1:]: tag.get(qn("w:val")) for tag in lvlTag.xpath("./*[@w:val]", namespaces=numbering_part.nsmap)} if "numFmt" not in style: numFmtVal = lvlTag.xpath("./mc:AlternateContent/mc:Fallback/w:numFmt/@w:val", namespaces=numbering_part.nsmap) if numFmtVal and numFmtVal[0] == "decimal": numFmt_format = lvlTag.xpath("./mc:AlternateContent/mc:Choice/w:numFmt/@w:format", namespaces=numbering_part.nsmap) if numFmt_format: style["numFmt"] = "decimal" + numFmt_format[0].split(",")[0] if style.get("numFmt") == "decimalZero": style["numFmt"] = "decimal01" abstractNumId2style[(int(abstractNumId), int(ilvl))] = style
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目前只发现这种基于decimal的格式，所以只针对这种自定义格式处理，其他类型的统一认为是没有自动编号。另外既然三位数的整数格式已经被我们命名为decimal001，那么也将二位数的decimalZero修改为decimal01。
目前测试出这个文件有以下这些numFmt：
bullet,cardinalText,chineseCounting,chineseLegalSimplified,decimal,decimalEnclosedCircleChinese,ideographTraditional,ideographZodiac,lowerLetter,lowerRoman,ordinal,ordinalText,upperLetter,upperRoman
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下面我们预先选择一些可能比较复杂的转换编写相应的函数：
正整数转换为大写字母
代码如下：
def int2upperLetter(num):
result = []
while num > 0:
num -= 1
remainder = num % 26
result.append(chr(remainder + ord('A')))
num //= 26
return "".join(reversed(result))
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运行
正整数转换为罗马数字
def int2upperRoman(num): t = [ (1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'), (90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I') ] roman_num = '' i = 0 while num > 0: val, syb = t[i] for _ in range(num // val): roman_num += syb num -= val i += 1 return roman_num
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运行
正整数转换为英文基数字
def int2cardinalText(num): if not isinstance(num, int) or num < 0 or num > 999999999999: raise ValueError( "Invalid number: must be a positive integer within four digits") base = ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] tens = ["", "", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] thousands = ["", "Thousand", "Million", "Billion"] def two_digits(n): if n < 20: return base[n] ten, unit = divmod(n, 10) if unit == 0: return f"{tens[ten]}" else: return f"{tens[ten]}-{base[unit]}" def three_digits(n): hundred, rest = divmod(n, 100) if hundred == 0: return two_digits(rest) result = f"{base[hundred]} hundred " if rest > 0: result += two_digits(rest) return result.strip() if num < 99: return two_digits(num) chunks = [] while num > 0: num, remainder = divmod(num, 1000) chunks.append(remainder) words = [] for i in range(len(chunks) - 1, -1, -1): if chunks[i] == 0: continue chunk_word = three_digits(chunks[i]) if thousands[i]: chunk_word += f" {thousands[i]}" words.append(chunk_word) words = " ".join(words).lower() return words[0].upper()+words[1:]
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运行
正整数转换为英文序数字
def int2ordinalText(num): if not isinstance(num, int) or num < 0 or num > 999999: raise ValueError( "Invalid number: must be a positive integer within four digits") base = ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] baseth = ['Zeroth', 'First', 'Second', 'Third', 'Fourth', 'Fifth', 'Sixth', 'Seventh', 'Eighth', 'Ninth', 'Tenth', 'Eleventh', 'Twelfth', 'Thirteenth', 'Fourteenth', 'Fifteenth', 'Sixteenth', 'Seventeenth', 'Eighteenth', 'Nineteenth', 'Twentieth'] tens = ["", "", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] tensth = ["", "", "Twentieth", "Thirtieth", "Fortieth", "Fiftieth", "Sixtieth", "Seventieth", "Eightieth", "Ninetieth"] def two_digits(n): if n <= 20: return baseth[n] ten, unit = divmod(n, 10) result = tensth[ten] if unit != 0: result = f"{tens[ten]}-{baseth[unit]}" return result thousand, num = divmod(num, 1000) result = [] if thousand > 0: if num == 0: return f"{int2cardinalText(thousand)} thousandth" result.append(f"{int2cardinalText(thousand)} thousand") hundred, num = divmod(num, 100) if hundred > 0: if num == 0: result.append(f"{base[hundred]} hundredth") return " ".join(result) result.append(f"{base[hundred]} hundred") result.append(two_digits(num)) result = " ".join(result).lower() return result[0].upper() + result[1:]
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运行
会复用前面的基数字转换规则。
正整数转换为中文数字
import re def int2Chinese(num, ch_num, units): if not (0 <= num <= 99999999): raise ValueError("仅支持小于一亿以内的正整数") def int2Chinese_in(num, ch_num, units): if not (0 <= num <= 9999): raise ValueError("仅支持小于一万以内的正整数") result = [ch_num[int(i)] + unit for i, unit in zip(reversed(str(num).zfill(4)), units)] result = "".join(reversed(result)) zero_char = ch_num[0] result = re.sub(f"(?:{zero_char}[{units}])+", zero_char, result) result = result.rstrip(units[0]) if result != zero_char: result = result.rstrip(zero_char) if result.lstrip(zero_char).startswith("一十"): result = result.replace("一", "") return result if num < 10000: result = int2Chinese_in(num, ch_num, units) else: left = num // 10000 right = num % 10000 result = int2Chinese_in(left, ch_num, units) + "万" + int2Chinese_in(right, ch_num, units) if result != ch_num[0]: result = result.strip(ch_num[0]) return result def int2ChineseCounting(num): return int2Chinese(num, ch_num='〇一二三四五六七八九', units='个十百千') def int2ChineseLegalSimplified(num): return int2Chinese(num, ch_num='零壹贰叁肆伍陆柒捌玖', units='个拾佰仟')
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运行
整体封装并改进
最终封装成为一个类：
import logging import re from io import BytesIO from PIL import Image from docx import Document, ImagePart from docx.oxml.ns import qn, nsmap from docx.text.paragraph import Paragraph from functools import lru_cache class WithNumberDocxReader: ideographTraditional = "甲乙丙丁戊己庚辛壬癸" ideographZodiac = "子丑寅卯辰巳午未申酉戌亥" def __init__(self, docx, gap_text="\t"): self.parts = [] self.docx = Document(docx) nsmap.update(self.docx.element.nsmap) self.numId2style = self.get_style_data() self.gap_text = gap_text self.cnt = {} self.cache = {} self.result = [] @property def texts(self): if self.result: return self.result.copy() self.clear() for paragraph in self.paragraphs: number_text = self.get_number_text(paragraph) line = number_text + paragraph.text.strip() if not line: continue self.result.append(line) return self.result.copy() def clear(self): self.result.clear() self.cnt.clear() self.cache.clear() @property @lru_cache def paragraphs(self): body = self.docx.element.body result = [] for p in body.xpath('w:p | w:sdt/w:sdtContent/w:p | w:p//v:textbox//w:p'): result.append(Paragraph(p, body)) return result @property def images(self): if self.parts: return self.parts.copy() related_parts = self.docx.part.related_parts for i, paragraph in enumerate(self.paragraphs, 1): for run in paragraph.runs: for drawing in run.element.drawing_lst: rid = drawing.xpath(".//a:blip/@r:embed") if not rid or rid[0] not in related_parts: continue part = related_parts[rid[0]] if isinstance(part, ImagePart): self.parts.append((i, part.partname, part.blob)) return self.parts def get_style_data(self): try: numbering_part = self.docx.part.numbering_part._element abstractId2numId = {num.abstractNumId.val: num.numId for num in numbering_part.num_lst} numId2style = {} for abstractNumIdTag in numbering_part.findall(qn("w:abstractNum")): abstractNumId = abstractNumIdTag.get(qn("w:abstractNumId")) numId = abstractId2numId[int(abstractNumId)] for lvlTag in abstractNumIdTag.findall(qn("w:lvl")): ilvl = lvlTag.get(qn("w:ilvl")) style = {tag.tag[tag.tag.rfind("}") + 1:]: tag.get(qn("w:val")) for tag in lvlTag.xpath("./*[@w:val]", namespaces=nsmap)} if "numFmt" not in style: numFmtVal = lvlTag.xpath("./mc:AlternateContent/mc:Fallback/w:numFmt/@w:val", namespaces=nsmap) if numFmtVal and numFmtVal[0] == "decimal": numFmt_format = lvlTag.xpath("./mc:AlternateContent/mc:Choice/w:numFmt/@w:format", namespaces=nsmap) if numFmt_format: style["numFmt"] = "decimal" + numFmt_format[0].split(",")[0] if style.get("numFmt") == "decimalZero": style["numFmt"] = "decimal01" numId2style[(numId, int(ilvl))] = style return numId2style except Exception as e: logging.warning("读取自动编号出错:" + e.__class__.__name__) @staticmethod def int2upperLetter(num): result = [] while num > 0: num -= 1 remainder = num % 26 result.append(chr(remainder + ord('A'))) num //= 26 return "".join(reversed(result)) @staticmethod def int2upperRoman(num): t = [ (1000, 'M'), (900, 'CM'), (500, 'D'), (400, 'CD'), (100, 'C'), (90, 'XC'), (50, 'L'), (40, 'XL'), (10, 'X'), (9, 'IX'), (5, 'V'), (4, 'IV'), (1, 'I') ] roman_num = '' i = 0 while num > 0: val, syb = t[i] for _ in range(num // val): roman_num += syb num -= val i += 1 return roman_num @staticmethod def int2cardinalText(num): if not isinstance(num, int) or num < 0 or num > 999999999: raise ValueError( "Invalid number: must be a positive integer within four digits") base = ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] tens = ["", "", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] thousands = ["", "Thousand", "Million", "Billion"] def two_digits(n): if n < 20: return base[n] ten, unit = divmod(n, 10) if unit == 0: return f"{tens[ten]}" else: return f"{tens[ten]}-{base[unit]}" def three_digits(n): hundred, rest = divmod(n, 100) if hundred == 0: return two_digits(rest) result = f"{base[hundred]} hundred " if rest > 0: result += two_digits(rest) return result.strip() if num < 99: return two_digits(num) chunks = [] while num > 0: num, remainder = divmod(num, 1000) chunks.append(remainder) words = [] for i in range(len(chunks) - 1, -1, -1): if chunks[i] == 0: continue chunk_word = three_digits(chunks[i]) if thousands[i]: chunk_word += f" {thousands[i]}" words.append(chunk_word) words = " ".join(words).lower() return words[0].upper() + words[1:] @staticmethod def int2ordinalText(num): if not isinstance(num, int) or num < 0 or num > 999999: raise ValueError( "Invalid number: must be a positive integer within four digits") base = ["Zero", "One", "Two", "Three", "Four", "Five", "Six", "Seven", "Eight", "Nine", "Ten", "Eleven", "Twelve", "Thirteen", "Fourteen", "Fifteen", "Sixteen", "Seventeen", "Eighteen", "Nineteen"] baseth = ['Zeroth', 'First', 'Second', 'Third', 'Fourth', 'Fifth', 'Sixth', 'Seventh', 'Eighth', 'Ninth', 'Tenth', 'Eleventh', 'Twelfth', 'Thirteenth', 'Fourteenth', 'Fifteenth', 'Sixteenth', 'Seventeenth', 'Eighteenth', 'Nineteenth', 'Twentieth'] tens = ["", "", "Twenty", "Thirty", "Fourty", "Fifty", "Sixty", "Seventy", "Eighty", "Ninety"] tensth = ["", "", "Twentieth", "Thirtieth", "Fortieth", "Fiftieth", "Sixtieth", "Seventieth", "Eightieth", "Ninetieth"] def two_digits(n): if n <= 20: return baseth[n] ten, unit = divmod(n, 10) result = tensth[ten] if unit != 0: result = f"{tens[ten]}-{baseth[unit]}" return result thousand, num = divmod(num, 1000) result = [] if thousand > 0: if num == 0: return f"{WithNumberDocxReader.int2cardinalText(thousand)} thousandth" result.append(f"{WithNumberDocxReader.int2cardinalText(thousand)} thousand") hundred, num = divmod(num, 100) if hundred > 0: if num == 0: result.append(f"{base[hundred]} hundredth") return " ".join(result) result.append(f"{base[hundred]} hundred") result.append(two_digits(num)) result = " ".join(result).lower() return result[0].upper() + result[1:] @staticmethod def int2Chinese(num, ch_num, units): if not (0 <= num <= 99999999): raise ValueError("仅支持小于一亿以内的正整数") def int2Chinese_in(num, ch_num, units): if not (0 <= num <= 9999): raise ValueError("仅支持小于一万以内的正整数") result = [ch_num[int(i)] + unit for i, unit in zip(reversed(str(num).zfill(4)), units)] result = "".join(reversed(result)) zero_char = ch_num[0] result = re.sub(f"(?:{zero_char}[{units}])+", zero_char, result) result = result.rstrip(units[0]) if result != zero_char: result = result.rstrip(zero_char) if result.lstrip(zero_char).startswith("一十"): result = result.replace("一", "") return result if num < 10000: result = int2Chinese_in(num, ch_num, units) else: left = num // 10000 right = num % 10000 result = int2Chinese_in(left, ch_num, units) + "万" + int2Chinese_in(right, ch_num, units) if result != ch_num[0]: result = result.strip(ch_num[0]) return result @staticmethod def int2ChineseCounting(num): return WithNumberDocxReader.int2Chinese(num, ch_num='〇一二三四五六七八九', units='个十百千') @staticmethod def int2ChineseLegalSimplified(num): return WithNumberDocxReader.int2Chinese(num, ch_num='零壹贰叁肆伍陆柒捌玖', units='个拾佰仟') def get_number_text(self, paragraph): if self.numId2style is None: return "" pr = paragraph._element.pPr if pr is None: return "" numpr = pr.numPr if numpr is None or numpr.numId.val == 0: return "" numId = numpr.numId.val ilvl = numpr.ilvl.val style = self.numId2style[(numId, ilvl)] numFmt: str = style.get("numFmt") lvlText = style.get("lvlText") isTxbxContent = paragraph._element.getparent().tag.endswith("txbxContent") for a, b, c in list(self.cnt.keys()): if a == numId and c == isTxbxContent and b > ilvl: del self.cnt[(a, b, c)] pos_key = (numId, ilvl, isTxbxContent) if pos_key in self.cnt: self.cnt[pos_key] += 1 else: self.cnt[pos_key] = int(style["start"]) pos = self.cnt[pos_key] num_text = str(pos) if numFmt.startswith('decimal'): num_text = num_text.zfill(numFmt.count("0") + 1) elif numFmt == 'upperRoman': num_text = self.int2upperRoman(pos) elif numFmt == 'lowerRoman': num_text = self.int2upperRoman(pos).lower() elif numFmt == 'upperLetter': num_text = self.int2upperLetter(pos) elif numFmt == 'lowerLetter': num_text = self.int2upperLetter(pos).lower() elif numFmt == 'ordinal': num_text = f"{pos}{'th' if 11 <= pos <= 13 else {1: 'st', 2: 'nd', 3: 'rd'}.get(pos % 10, 'th')}" elif numFmt == 'cardinalText': num_text = self.int2cardinalText(pos) elif numFmt == 'ordinalText': num_text = self.int2ordinalText(pos) elif numFmt == 'ideographTraditional': if 1 <= pos <= 10: num_text = self.ideographTraditional[pos - 1] elif numFmt == 'ideographZodiac': if 1 <= pos <= 12: num_text = self.ideographZodiac[pos - 1] elif numFmt == 'chineseCounting': num_text = self.int2ChineseCounting(pos) elif numFmt == 'chineseLegalSimplified': num_text = self.int2ChineseLegalSimplified(pos) elif numFmt == 'decimalEnclosedCircleChinese': pass self.cache[pos_key] = num_text for i in range(0, ilvl + 1): lvlText = lvlText.replace(f'%{i + 1}', self.cache.get((numId, i, isTxbxContent), "")) suff_text = {"space": " ", "nothing": ""}.get(style.get("suff"), self.gap_text) lvlText += suff_text return lvlText if __name__ == '__main__': doc = WithNumberDocxReader(r"编号测试1.docx", " ") for text in doc.texts: print(text) for i, name, image_bytes in doc.images: print(i, name) image = Image.open(BytesIO(image_bytes)) image.show()
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调用测试：
if __name__ == '__main__':
doc = WithNumberDocxReader(r"编号测试2.docx", "")
for text in doc.texts:
print(text)
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顺利达到打印出对应的字符：
点符 1.十进制数 01.零加十进制数 001 零零加十进制数 0001 零零零加十进制数 I 大写罗马数字 (I) II 大写罗马数字 (II) i 小写罗马数字 A.大写字母A a 小写字母 (a) 0th 序数 (1st, 2nd, 3rd) Twelve 基数字 (One, Two Three) First 序数字 (First, Second, Third) 癸 甲乙丙丁戊己庚辛壬癸 壹 中文大写数字 10 圆圈数字 子 子丑寅卯辰巳午未申酉戌亥 第一章 中文数字
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