sys.argv is not taking provided argument in python 3

Question

I'm using python 3.3.0 in windows 8.

This is the code which I'm trying to deal with:-

import sys

for arg in sys.argv:
    if arg == "-u":
        url = sys.argv[2]
        print (url)
        sys.exit(1)

So, When I run my program in cmd like:

python my.py -u http://www.aytelecom.ae/news-details.php?id=5&type=2

I'm getting only 'http://www.aytelecom.ae/news-details.php?id=5' as my sys.argv[2] or in url variable!

So, can anybody tell me that why it's skipping '&type=2'? Am I doing mistake anywhere or what's wrong with python 3?

seq.index(x) returns the index of the first occurrence of x in seq or raises ValueError if x is not in seq. You could use it instead of the for-loop. — jfs
– jfs, Commented Oct 24, 2012 at 13:12
Yup, I got your point but I think it would be useful for only one parameter. What if I have more options to check for? And when I tried to run this: import sys sys.argv.index("-u") url = sys.argv[2] print ("URL:", url) sys.exit(1) It gives something in the last line like: The system cannot find the file specified. — magneto
– magneto, Commented Oct 24, 2012 at 13:27

jfs · Accepted Answer · 2012-10-24 12:53:02Z

6

& is a metacharacter in shell. Put the url in quotes.

answered Oct 24, 2012 at 12:53

jfs

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1 Comment

oh my god, it was so easy and I have wasted whole day on Google to get a solution! Anyway, Thanks for your quick response. It's really a nice site.