CODEWITHSUNDEEP
python
RonicK
RonicK

Reputation: 229

Parse path as argument in python script

This is a part of python script xyz.py.

#!/usr/bin/env python

from openpyxl import Workbook
import os
wb = Workbook()
path = "/home/Final_analysis/"
#print(os.listdir())
lis = os.listdir(path)

I wanted to use this script with different input files which are stored in different directories. For this script to run on multiple files I would have to change this part of the script every time.

path= "/home/Final_analysis/"

Can I parse the path as argument after python script for example

xyz.py path_to_my_file

in the command so that each time I don't have to change the script?

I tried running script as above after writing sys.argv[1] in path 

#!/usr/bin/env python

from openpyxl import Workbook
import os
wb = Workbook()
path = "sys.argv[1]"
#print(os.listdir())
lis = os.listdir(path)

and ran command xyz.py /home/final_analysis

but it still doesn't detect the path. I am getting following error

Traceback (most recent call last):
  File "combine_excel.py", line 8, in <module>
    lis = os.listdir(path)
OSError: [Errno 2] No such file or directory: 'sys.argv[1]'

I am using in python 2.7.6.

Upvotes: 0

Views: 391

Answers (2)

Max
Max

Reputation: 1175

In addition to what people have commented on you post, you can also use the "{}".format() syntax. A really clean way to insert string into an already formed string.

Upvotes: 0

9769953
9769953

Reputation: 12201

Don't pass sys.argv[1] as a string, use it directly: path = sys.argv[1]

#!/usr/bin/env python

from openpyxl import Workbook
import os
import sys
wb = Workbook()
path = sys.argv[1]
#print(os.listdir())
lis = os.listdir(path)

(per zwer's comment to the question.)

Upvotes: 1

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