使用Python启动Excel文件 [英] Use Python to launch Excel file

问题描述

当我尝试 os.system("open " + 'myfile.xlsx')
我得到输出'0'

when i try os.system("open " + 'myfile.xlsx')
i get the output '0'

类似地，尝试
os.system("start excel.exe myfilepath")
给出结果32512

similarly, trying
os.system("start excel.exe myfilepath")
gives the result 32512

我已经导入了操作系统和系统，并且在Mac上.如何更改此设置，以便它确实启动该excel文件?出于好奇，它打印出的数字是什么意思?

I have imported os and system, and I'm on mac. How can I change this so it does actually launch that excel file? And out of curiosity, what do the numbers it prints out mean?

谢谢！

推荐答案

如果只想打开excel应用程序，则可以使用子过程:

If you only want to open the excel application you could use subprocess:

import subprocess
subprocess.check_call(['open', '-a', 'Microsoft Excel'])

您还可以使用os并打开特定文件:

You can also use os and open a specific file:

import os
os.system("open -a 'path/Microsoft Excel.app' 'path/file.xlsx'")

如果您想在python中打开一个excel文件并对其进行修改，则有许多软件包可以用作 xlsxwriter ， xlutils 和 openpyxl 我更喜欢后者.

If you on other hand want to open an excel file within python and modify it there's a number of packages to use as xlsxwriter, xlutils and openpyxl where the latter is prefered by me.

另一个说明，如果您使用的是Mac，则excel应用程序不是.exe

Another note, if you're on mac the excel application isn't .exe

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