openpyxl：在没有 zipfile 的情况下获取工作表的 xml 源代码

Question

from openpyxl import load_workbook

wb = load_workbook('file.xlsx')
ws = wb['Sheet1']

有没有办法检索表示

ws

对象的xml代码？ 注意：我想避免使用

zipfile

模块。相反，我试图从 ws.

 中提取 xml 

directly

我阅读了

openpyxl

源代码并且正在玩

lxml

- 但没有成功。

Answer 1

我自己想通了。您可以在保存工作簿时捕获它，而不是通过解压缩已保存的工作簿来提取 xml。

wb.save

方法使用了

ExcelWriter

类，我为此目的修改了该类：

import openpyxl
from openpyxl.writer.excel import *

class MyExcelWriter(openpyxl.writer.excel.ExcelWriter):
    def write_worksheet(self, ws):
        ws._drawing = SpreadsheetDrawing()
        ws._drawing.charts = ws._charts
        ws._drawing.images = ws._images
        if self.workbook.write_only:
            if not ws.closed:
                ws.close()
            writer = ws._writer
        else:
            writer = WorksheetWriter(ws)
            writer.write()
        ws._rels = writer._rels
        
        # my addition starts here
        if ws.title == 'My Sheet':
            with open(writer.out, 'r', encoding='utf8') as file:
                xml_code = file.read()

            # my code continues here...
        # my addition ends here

        self._archive.write(writer.out, ws.path[1:])
        self.manifest.append(ws)
        writer.cleanup()

openpyxl.writer.excel.ExcelWriter = MyExcelWriter

write_worksheet

函数创建一个临时xml文件，其路径存放在

writer.out

.
请记住，

from <module> import *

被认为是不好的做法——谨慎使用。