我有一个excel文件列表与最后一行相似。它包含有关客户的私人信息(他的姓名,电话)。每个excel文件对应一个客户端。我需要为每个客户端创建一个包含所有数据的excel文件。我决定自动做到这一点,所以看到openpyxl
图书馆。我写了下面的代码,但它不能正常工作。无法读取excel文件,使用openpyxl
import openpyxl
import os
import glob
from openpyxl import load_workbook
from openpyxl import Workbook
import openpyxl.styles
from openpyxl.cell import get_column_letter
path_kit = 'prize_input/kit'
#creating single document
prize_info = Workbook()
prize_sheet = prize_info.active
file_array_reciever = []
for file in glob.glob(os.path.join(path_kit, '*.xlsx')):
file_array_reciever.append(file)
row_num = 1
for f in file_array_reciever:
f1 = load_workbook(filename=f)
sheet = f1.active
for col_num in range (3, sheet.max_column):
prize_sheet.cell(row=row_num, column=col_num).value = \
sheet.cell(row=sheet.max_row, column=col_num).value
prize_info.save("Ex.xlsx")
我得到这个错误:
Traceback (most recent call last):
File "/Users/zkid18/PycharmProjects/untitled/excel_test.py", line 43, in <module>
f1 = load_workbook(filename=f)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/site-packages/openpyxl/reader/excel.py", line 183, in load_workbook
wb.active = read_workbook_settings(archive.read(ARC_WORKBOOK)) or 0
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/zipfile.py", line 1229, in read
with self.open(name, "r", pwd) as fp:
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/zipfile.py", line 1252, in open
zinfo = self.getinfo(name)
File "/Library/Frameworks/Python.framework/Versions/3.5/lib/python3.5/zipfile.py", line 1196, in getinfo
'There is no item named %r in the archive' % name)
KeyError: "There is no item named 'xl/workbook.xml' in the archive"
看起来它是读文件有问题。
我不明白它在归档中获取名为'xl/workbook.xml'
的项目的位置。
也许其中一个excel文件已损坏或使用不受openpyxl支持的功能。 – Muposat
确保文件路径正确,您可以在每次循环迭代中打印出'f'来确认。 – schaazzz
你试图阅读的文件几乎肯定存在问题。 –