i have some problem with if statement(Python)

Question

I was working on reading some data from excel. However, i have countered a simple if statement. Not quite sure what have i missed. When i run my code, it will not go into the if statement but it will instead jump to else statement even though the value specified is exactly the same. below is my code. Thank you in advance!

from openpyxl import load_workbook
wb = load_workbook()
sheet_ranges = wb['Sheet1']
xd_date_count = 3
cookie = sheet_ranges['A' + str(xd_date_count)].value # i have a value "None" stored in A3 in this excel
print(cookie) # when i print this, it prints out None
if cookie == "None":
    print("ok")
else:
    print("no")

Are you sure that your cookie is "None" and not None? The former would happen for a cell with the literal text "None", whereas the latter would happen for an empty cell, probably — Green Cloak Guy
– Green Cloak Guy, Commented Apr 7, 2021 at 15:06
try printing print(repr(cookie), repr("None")) to check what Green Cloak Guy suggested — CallMeStag
– CallMeStag, Commented Apr 7, 2021 at 15:06

David · Accepted Answer · 2021-04-07 15:07:02Z

1

If you are looking for a None object, you should try None in stead of "None", as this will be interpreted as a string.

In addition

if cookie is None:

is nicer to read, but that's just a personal flavor thing

edited Apr 7, 2021 at 15:07

answered Apr 7, 2021 at 15:06

David

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3 Comments

MegaIng Over a year ago

'Just a wild guess' shouldn't be an answer. This is more a comment.

David Over a year ago

Agree. Changed the text.

David Over a year ago

With the brackets around the word None, you where literally looking for the string "None", while this is a None object which is not a string and therefore be checked using None.

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