Python OSErrror errno22 when creating a file

Question

I get errno 22 with this code:

import datetime
now = str(datetime.datetime.now())
filename = "log_{}".format(now)
logFile = open(filename, "w+")

probably related https://stackoverflow.com/questions/25584124/oserror-errno-22-invalid-argument-when-use-open-in-python#25584140 — Martin Gergov, Sep 13 '18 at 16:39

score 0 · Accepted Answer · answered Sep 13 '18 at 16:42

0

Quick fix if you are okay with : being replaced by .:

import datetime
now = str(datetime.datetime.now())
filename = "log_{}".format(now).replace(':', '.')
filename
# 'log_2018-09-13 11.39.42.216000'
logFile = open(filename, "w+")

It is an issue with the filename containing ":" characters.

answered Sep 13 '18 at 16:42

d_kennetz

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thanks! now it works! – DanieleManna Sep 13 '18 at 16:43
I the past I've just used `;` it looks about the same, the problem with replacing `:` with `.` is when you go to open the file and delimit the timestamp you can't separate the `.` for microseconds as easily. – tgikal Sep 13 '18 at 16:43

Python OSErrror errno22 when creating a file

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