Hi, you could save yourself some hassle and do
minipstr = '1.0.0.1' maxipstr = '1.0.15.16'
minip = map(int, minipstr.split('.')) maxip = map(int, maxipstr.split('.'))
iplist = []
for a in range(minip[2], maxip[2]+1):
... if a < maxip[2]: ... for b in range(minip[3], 255): ... iplist.append('.'.join(map(str, [minip[0],minip[1], a, b]))) ... else: ... for b in range(minip[3], minip[3]+1): ... iplist.append('.'.join(map(str, [minip[0],minip[1], a, b])))
Eek, that's a bit Perlish, might want to break the iplist.append line into
ipintlist = [minip[0],minip[1], a, b] ipstrlist = map(str, ipintlist) iplist.append('.'.join(ipstrlist))
I don't think this will work correctly with for example minipstr = '1.0.0.1' maxipstr = '1.2.0.0'
I would break this problem up conceptually. It has maybe four different parts:
- convert a string representation of an IP address to a useful representation. Liam shows how to do this above, a list of ints is easy to work with.
- convert the useful representation back to a string. Again, Liam shows how to do this.
- compare two IPs. If the IPs are represented as lists of ints, you can compare them directly:
>>> a=[1,2,1,5]
>>> b=[1,2,3,4]
>>> a>b
False
>>> a<b
True
>>> a==[1,2,1,5]
True
- increment an IP. This is the hardest part. You have to implement a counter that works with the list representation. Here is one way to do it - this function does the right thing with the last 'digit', then if there was a carry it calls itself recursively to increment the next digit. It rolls over from [255, 255, 255, 255] to [0, 0, 0, 0]:
def incr(n, limit, ix=None):
''' Increment a number base (limit+1) represented as a list of ints '''
if ix is None: # initial call starts at the end
ix = len(n) - 1
if ix < 0: # Off the end, give up
return
if n[ix] < limit: # Normal increment
n[ix] += 1
else: # Increment with carry
n[ix] = 0
incr(n, limit, ix-1)
a=[1,2,1,5] incr(a, 255) print a
a=[1,2,1,255] incr(a, 255) print a
a=[1,2,255,255] incr(a, 255) print a
a=[1,255,255,255] incr(a, 255) print a
a=[255,255, 255,255] incr(a, 255) print a
## prints [1, 2, 1, 6] [1, 2, 2, 0] [1, 3, 0, 0] [2, 0, 0, 0] [0, 0, 0, 0]
Kent
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