Question

这是一个非常愚蠢的问题，但我试图学习python，并且我在使用xlrd读取Excel文件时遇到了困难。我在网上找到了这个脚本，但我无法弄清楚我应该在哪里填充我的文件名以使其打开。

from future import print_function
from os.path import join, dirname, abspath, isfile
from collections import Counter
import xlrd
from xlrd.sheet import ctype_text   

def get_excel_sheet_object(fname, idx=0):
    if not isfile(fname):
        print ('File doesn't exist: ', fname)
            # Open the workbook and 1st sheet
    xl_workbook = xlrd.open_workbook(fname)
    xl_sheet = xl_workbook.sheet_by_index(0)
    print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

return xl_sheetsdf

Answer 1

from future import print_function
from os.path import join, dirname, abspath, isfile
from collections import Counter
import xlrd
from xlrd.sheet import ctype_text   


def get_excel_sheet_object(fname, idx=0):
    if not isfile(fname):
        print ('File doesn't exist: ', fname)
            # Open the workbook and 1st sheet
    xl_workbook = xlrd.open_workbook(fname)
    xl_sheet = xl_workbook.sheet_by_index(0)
    print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

    return xl_sheet

xl_sheet_obj = get_excel_sheet_object('FILE_NAME_HERE')

在此对象为Excel工作表对象之后，对xl_sheet_obj执行任何操作。

Answer 2

您可以输入文件名：

调用函数时

get_excel_sheet_object（ “myfile.xlsx”）
```
       OR
```
fname =“myfile.xlsx”

get_excel_sheet_object（FNAME）

您的计划中的Raw：

def get_excel_sheet_object(idx=0):

      fname = "myfile.xlsx"

      if not isfile(fname):
         print ("File doesn't exist: ", fname)
      # Open the workbook and 1st sheet
      xl_workbook = xlrd.open_workbook(fname)
      xl_sheet = xl_workbook.sheet_by_index(0)
      print (40 * '-' + 'nRetrieved worksheet: %s' % xl_sheet.name)

      return xl_sheet

在python脚本中指定filename以打开Excel工作簿

2 个答案: