How to extract the hyperlink cell from excel sheet using python

Question

I am trying to open the hyperlinks of the individual cells in excel spreadsheet.This is the code i tried

import xlrd
import logging
wb=xlrd.open_workbook("hhh.xls")
sh=wb.sheet_by_index(3)
wsname=[]
for row in range(sh.nrows):
    if sh.cell_value(row,15)=="Table":
        print sh.cell_value(row,15)
        print sh.cell_value(row,2)

        link =sh.hyperlink_map.get(row,2)
        url='(No URL)' if link is None else link.url_or_path
        wsname = link.textmark.split('!')[0]
        logging.debug("link=%s wsname=%s" % (link.textmark, wsname))
        wstable=wb.sheet_by_name(wsname)

Please help me the problem this is the error

Traceback (most recent call last): File "reading.py", line 12, in url='(No URL)' if link is None else link.url_or_path AttributeError: 'int' object has no attribute 'url_or_path'

help........

Answer 1

From the documentation, you need to pass in a tuple:

link = sh.hyperlink_map.get((row,2))