Comments (4)
Hi thank you for raising this issue.
Please provide us the line where your code breaks and the 5 lines of code before that so we can investigate if it's a problem with your code or with gspread.
from gspread.
Here is the complete code
import gspread
def sheet_api():
gc = gspread.service_account(filename='./config/service_account.json')
# public spreadsheet - all three references to same spreadsheet
try:
sh = gc.open("https://docs.google.com/spreadsheets/d/1gGF4o1C-ytu0aaCdVm6D73wGwPCMvETpqsScTXQzAPM/edit#gid=1164501076")
print(sh.sheet1.get('A1'))
except gspread.exceptions.SpreadsheetNotFound:
pass
try:
sh = gc.open("1gGF4o1C-ytu0aaCdVm6D73wGwPCMvETpqsScTXQzAPM")
print(sh.sheet1.get('A1'))
except gspread.exceptions.SpreadsheetNotFound:
pass
try:
sh = gc.open("CheckWatt serial numbers")
print(sh.sheet1.get('A1'))
except gspread.exceptions.SpreadsheetNotFound:
pass
# private spreadsheet works
sh = gc.open("torsdag")
print(sh.sheet1.get('A1'))
# python3 sheet.py
def main():
sheet_api()
if __name__ == "__main__":
main()
from gspread.
Hi thank you for sharing.
The issue is located here:
sh = gc.open("https://docs.google.com/spreadsheets/d/1gGF4o1C-ytu0aaCdVm6D73wGwPCMvETpqsScTXQzAPM/edit#gid=1164501076")
As mentioned in the documentation here
The method open()
takes a parameter title
which is the title of the spreadsheet.
It then searches through all you private + shared filed for the file matching that name.
If you wish to open a public spreadsheet file and you don't own it, please use the open_by_key()
method, see example here
from gspread.
Confirmed! open_by_key()
is the answer! Thanks for the quick response!
from gspread.
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from gspread.