Calling functions based on a if condition in python

Question

i have to call the main function based on the number of arguments passed. When i call the script the functions are not working.

Sample Code:

    def func1():
      somecode
    def func2():
      somecode
    def func3():
      somecode
    def main():
      if len(sys.argv) == "5":
        func1()
        func3()
      elif len(sys.argv) == "8":
       func2()
       func3()
    if __name__ == '__main__':
      main()

What is your question? On a side note, one obvious problem: don't compare strings ("5") and integers (len(...)). — Anthony Labarre
– Anthony Labarre, Commented Feb 11, 2018 at 15:28
when i call the script, the func1 and func2 and func3 are not working. — Ramesh Eol
– Ramesh Eol, Commented Feb 11, 2018 at 15:30
Possible duplicate? stackoverflow.com/questions/40403108/… — mkrieger1
– mkrieger1, Commented Feb 11, 2018 at 15:31
5 != "5", as mentioned, use if len(sys.argv) == 5: and elif len(sys.argv) == 8: to correctly compare the number of arguments passed. — eugenhu
– eugenhu, Commented Feb 11, 2018 at 15:35

FlyingTeller · Accepted Answer · 2018-02-11 15:36:08Z

0

In your code you are comparing len(sys.argv) with a string:

  if len(sys.argv) == "5":
    func1()
    func3()
  elif len(sys.argv) == "8":
   func2()
   func3()

changing to

  if len(sys.argv) == 5:
    func1()
    func3()
  elif len(sys.argv) == 8:
   func2()
   func3()

should do the trick

answered Feb 11, 2018 at 15:36

FlyingTeller

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Comments

BoarGules · Accepted Answer · 2018-02-11 15:36:46Z

0

Your code is not calling those functions because this if-test:

if len(sys.argv) == "5":

is always False. The function len() returns an integer and an integer in Python is never equal to a string. Do this instead:

if len(sys.argv) == 5:

answered Feb 11, 2018 at 15:36

BoarGules

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