On 11/26/2016 12:55 PM, Dennis Lee Bieber wrote:
On Sat, 26 Nov 2016 08:12:46 -0800 (PST), junko...@gmail.com declaimed the
following:
import csv
with open('\\192.168.0.1\fe18cb0618cabd41\ninjatrader$EURUSDTestRun
2016-11-25-11-11.csv','r') as f:
reader=csv.reader(f)
for row in reader:
print(row)
File "C:/Users/Suresh/PycharmProjects/untitled/test.py", line 2, in <module>
with open('\\192.168.0.1\fe18cb0618cabd41\ninjatrader$EURUSDTestRun
2016-11-25-11-11.csv','r') as f:
OSError: [Errno 22] Invalid argument:
'\\192.168.0.1\x0ce18cb0618cabd41\ninjatrader$EURUSDTestRun
2016-11-25-11-11.csv'
First suggestion... Make it a RAW string. What you have has an embedded
new-line character making it:
\\192.168.0.1\fe18cb0618cabd41
injatrader$EURUSDTestRun 2016-11-25-11-11.csv
Or use forward slashes:
with open('/192.168.0.1/fe18cb0618cabd41/ninjatrader$EURUSDTestRun
2016-11-25-11-11.csv','r') as f:
Second... does Python open() accept web addresses? What is at
192.168.0.1 that you don't have some more sensible path to it? 192.168.0.1
is a private local network address. You are on a Windows machine so if the
file is available on a local network node you can probably mount that node
(Computer/Map Network Drive) and access the file as something like
Z:/ninjatrader...
--
Terry Jan Reedy
--
https://mail.python.org/mailman/listinfo/python-list
