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脚本之家。
