实现按目录拆分工作簿，源数据如下图

按目录拆分成N个文件。

上代码，没有找是否有整个sheet 复制的，先逐个cell复制解决问题。：

# encoding: utf-8"""@author: 陈年椰子@contact: hndm@qq.com@version: 1.0@file: Split_Xls.py@time: 2019/9/24 0028 15:04说明"""def Split_Xls(xls_file): from openpyxl import load_workbook from openpyxl import Workbook wb = load_workbook(xls_file) sheet_list = wb.sheetnames print(sheet_list) a_sheet = wb['目录'] for i in range(3,6): sheet_name = a_sheet['B{}'.format(i)].value if sheet_name is None: break if sheet_name == '': break sr_sheet = wb[sheet_name] new_file_name = "{}.xlsx".format(sheet_name) print(sheet_name) wb_tg = Workbook() ws = wb_tg.active ws.title = sheet_name # 两个for循环遍历整个excel的单元格内容 for i, row in enumerate(sr_sheet.iter_rows()): for j, cell in enumerate(row): # print(i,j,cell.value) ws.cell(row=i + 1, column=j + 1, value=cell.value) wb_tg.save(new_file_name) wb_tg.close() wb.close() def Split_Xls2(xls_file): # 这个是通过删除其他的工作表，只留下要保存的工作表，这样就可以整个表复制，包括样式，过程曲折，但能达到效果。 from openpyxl import load_workbook wb = load_workbook(xls_file) sheet_list = wb.sheetnames print(sheet_list) work_list = [] a_sheet = wb['目录'] for i in range(3,6): sheet_name = a_sheet['B{}'.format(i)].value if sheet_name is None: break if sheet_name == '': break work_list.append(sheet_name) wb.close() for sheet_name in work_list: new_file_name = "{}.xlsx".format(sheet_name) print('处理工作表', sheet_name, '\t保存文件', new_file_name) wb = load_workbook(xls_file) # print(wb.sheetnames) for del_sheet in sheet_list: if del_sheet != sheet_name: # print('del',del_sheet) wb.remove(wb[del_sheet]) wb.save(new_file_name) wb.close() Split_Xls2('test.xlsx')

以上就是本文的全部内容，希望对大家的学习有所帮助，也希望大家多多支持。