CODEWITHSUNDEEP
pythoncommand-line-arguments
Vipul
Vipul

Reputation: 125

Not recognizing command line argument in Python

I am trying to add a command line argument depending on some values returned by the functions. When I am not giving that argument it says:

main.py: error: argument -opp/--operator is required

When I am giving the argument it says:

main.py: error: unrecognized arguments: -opp +

Following is the piece of EDITED code (as told in one of the answers):

parser.add_argument('-z', help='Help msg', required=True)
args, unknown = parser.parse_known_args()
value = some_functions(args.z)
if value == some_particular_value:
    parser.add_argument('-opp','--operator',help='Some help msg',required=True)
args = parser.parse_args()

Please help me in adding this argument. Thanks!

Upvotes: 0

Views: 3481

Answers (1)

shybovycha
shybovycha

Reputation: 12225

There are, though, a couple of mistakes in your code. Here's the corrected version:

import argparse

parser = argparse.ArgumentParser()
parser.add_argument('-z', help = 'Help msg', required = True)

args, _ = parser.parse_known_args()

# value = some_functions(args.z)

if value == some_particular_value:
    parser.add_argument('-opp', '--operator', help = 'Some help msg', required = True)

    # args2, _ = parser.parse_known_args()
    # some_function2(args2.operator)

So, let's analyse your mistakes:

Assigning instead of comparing

That's typical newbie mistake. Within a conditional operator (if, case...) you set the value instead of checking it. The difference is in amount of the = sign.

If you assign value, the condition in the operator will be always True and test will always succeed (in most of programming languages and cases).

Check this out:

a = 1

if a = 2:
  print a

This may print 2 in some languages (like C or Java; using the correct syntax). Why? You've just set it! Yet, Python is smart enough to tell you about your mistake:

File "<stdin>", line 1
    if a = 2:
         ^
SyntaxError: invalid syntax

And compare it to this:

a = 1

if a == 2:
  print a

This will not print anything. Because the if test did not pass.

Assigning value instead of calling method

You want to be using the method add_argument instead of re-defining the parser variable, right?

parser = add_argument(...)

That's something like I've described above. You should be calling a method of a parser variable, not defining its new value:

parser.add_argument(...)

Re-parsing arguments is missing?

You did not show the part of the code where you check for the operator argument. Note: you should parse your arguments again, when defined a new argument:

parser.add_argument(...)
args, _ = parser.parse_known_arguments()

Then you will get a new argument in the args variable.

Using the wrong name of argument?

Again, you are missing part of code, where you check for the operator argument' value. If you are trying to access it with

args.opp # whoops...

Then you'd just get an error saying There's no argument 'opp'!, because it has its full name and should be accessed with it:

args.operator # aaah, here it is!

Upvotes: 2

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