While loop with sys.argv

[huskeraider]

I'm just learning python. I'm trying to figure out while loops with s.argv in a larger script on my network. hopefully this small example will show basically what i'm trying to do.

argtest.py

import sys

for arg in sys.argv:
print arg
#which gives me:
Life
is
good
but
could
be
better

using a while loop i'm trying to print out the last 4 arguments:
"but could be better", while allowing any additional arguments i add to be printed (8,9,10,so on)

I'm trying to do this with a list of servers in my network but this a shorter example.

I'm sure most of you can do this in your sleep but i'm just learning and it's not so easy for me.

I would greatly appreciate any assistance.
thanks.

[lllllIllIlllI]

So basically do you always want the last 4 arguments printed out? Because if that is the case i wouldn't use a while loop i would go something like this:

import sys

#-4 means start from the back and count back 4
#and then the rest of the arguments (:
for arg in sys.argv[-4:]:
    print arg

Hope that helps ;)

[huskeraider]

So basically do you always want the last 4 arguments printed out? Because if that is the case i wouldn't use a while loop i would go something like this:

import sys

#-4 means start from the back and count back 4
#and then the rest of the arguments (:
for arg in sys.argv[-4:]:
    print arg

Hope that helps ;)

i would prefer to the example you provided but i think part of the deal is me learning while loops. I've seen several examples of while loops but couldn't make it work with the above criteria.

Thanks,

[zachabesh]

From what you've said, it sounds like you have to print out elements 3 through 6 (if 0 is the first element) of a list. This will print those 4 elements, doesn't matter how long the list is.

n = 3
while n <= 6:
   print lst[n]
   n += 1

You could also use a for loop:

for x in range(3,7):
   print lst[x]

[huskeraider]

So, i would replace "n" the 3rd argument?

thanks for the help.

[lllllIllIlllI]

Actaully if you wanted the third item till the sixth item, the above code would almost be correct. Apart from one thing:

#n = 3
#if n = 3, then the fourth item will be printed 
n=2

while n <= 6:
   print lst[n]
   n += 1

This is because a list counts from 0, then 1,2,3 and so on. So when you go list[2] you're actually asking for the 3rd element in the list.

Hope that helps :)