Open a Python file with command line arguments

Question

NOTE : I don't know if this has been asked before, but I couldn't find what wording would be used for asking this question

I am creating a python file, and inside it needs to have a name of a file. What I want is when you put python Compiler.py <Filename> in the command line, it takes the filename and uses it in the python file.

How would I go about doing this?

Do you want compile a python code?

Saeed Bolhasani
– Saeed Bolhasani

2018-05-29 17:41:51 +00:00
Commented May 29, 2018 at 17:41 — Saeed Bolhasani
– Saeed Bolhasani, Commented May 29, 2018 at 17:41
@SaeedBolhasani Python is interpreted, not compiled.

Ctrl S
– Ctrl S

2018-09-21 18:47:03 +00:00
Commented Sep 21, 2018 at 18:47 — Ctrl S
– Ctrl S, Commented Sep 21, 2018 at 18:47

Ctrl S · Accepted Answer · 2018-05-29 18:06:02Z

5

someVariable = sys.argv[1]

This will take the first element after the file name and store it in the variable.

someOtherVariable = sys.argv[2]

will take the second element, and so forth.

python Compiler.py firstArg secondArg NthArg

Note: You will need to import sys to use the sys module.

edited May 29, 2018 at 18:06

answered May 29, 2018 at 17:42

Ctrl S

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2 Comments

MadPapo Over a year ago

The terminology to search on google is "how to pass script/process input arguments in python".They are managed in all programming languages

Ctrl S Over a year ago

@MadPapo This would be more appropriate as a comment on the question.

Ryan Schaefer · Accepted Answer · 2018-05-29 17:42:57Z

1

Use sys.argv to get the args that are passed to the script.

import sys
open(sys.argv[1], 'w').write('hello world')

answered May 29, 2018 at 17:42

Ryan Schaefer

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