Why does Python seem to run the else statement although the if clause is true?-CodePudding

I tried to write a simple program in Python 3.10 that opens (at least) one browser window and can take input from the command line to open additional windows:

import webbrowser
import sys

if len(sys.argv) > 1:
    for a in range(len(sys.argv)):
        webbrowser.open(sys.argv[a])
else:
    webbrowser.open("www.google.com")

I run that from a one-line batchfile so I can enter additional websites to open. That does work well, however, it always opens Google, no matter if command line arguments are passed or not (if I just run it without arguments, it opens Google. If I run it with e.g. LinkedIn as an argument, it opens Google and LinkedIn. If I use wikipedia and LinkedIn as arguments, it opens both of these plus Google etc).

However what I want is that Google is only opened when no arguments are passed... The strange thing is that when I run the exact same code, but replace the webbrowser.open() with a print statement, it does exactly what it is supposed to do:

import sys

if len(sys.argv) > 1:
    for a in range(len(sys.argv)):
        print(sys.argv[a])
else:
    print("www.google.com")

Passing command line arguments to this code causes it to only print the arguments, "www.google.com" is only printed when no arguments are passed.

I am sorry, but I am completely lost why these two codes act differently. I address both by running a batch file that goes

@py.exe C:\Users\[Route to the file]\[File].py %*

What I have tried is replacing the else statement by elif len(sys.argv) == 1, but that acts exactly the same. The only differnce I see is the usage of the webbrowser module. Can importing a module change behaviour of the code in that way?

Thank you in advance!

CodePudding user response：

argv captures the python script name and the website, which you already know because you're checing if len > 1. However you're then iterating over all args, causing the browser to attempt to load both the script name and the website as websites. Since the script name is not a valid url, I'm assuming your browser is opening to your default page, which happens to be google.com.

You need to skip the first arg, which you can do by sys.arvg[1:]

import webbrowser
import sys

if len(sys.argv) > 1:
    for a in sys.argv[1:]:
        webbrowser.open(a)
else:
    webbrowser.open("www.google.com")