Testing for Null?

c0mrade jacksingleton1 at gmail.com
Sun Jun 29 03:12:35 CEST 2008

Try something like this...

list = ['lkdfjsldk', None, '', '0', 'slfkjsdlfj', 'lsdgjdlfg', False, True]
for n, it in enumerate(list):
    if not it: print 'Error on this definition'
    else: print '%d. %s' % (n+1, it)

1. lkdfjsldk
Error on this definition
Error on this definition
4. 0
5. slfkjsdlfj
6. lsdgjdlfg
Error on this definition
8. True

Alexnb wrote:
> I am having a problem with a list value that is empty. I have a list of
> definitions called mainList. the 5th value in the list doesn't have
> anything
> in it. In this case, the values are definitions; also, in this case just
> the
> word cheese is defined. Here is my output to the console:
> 5.   a sprawling,weedy plant having small lavender or white flowers and
> round, flat, segmented fruits thought to resemble little wheels of cheese.
> 6.
> 7.  an ingot or billet made into a convex, circular form by blows at the
> ends.
> I've made it so where the numbers, the period, and two spaces follow that,
> then the definition. However, as you can see in 6, there is nothing. Here
> is
> the code to print all this:
> n=0
> for x in mainList:
>     if mainList[n] == "":
>         print "Error on this definition"
>     else:
>         print str(n+1)+".  "+str(mainList[n])
>     n=n+1
> Now the two "" is where I need to figure out if it is empty. What is up
> right now doesn't work; or at least doesn't give the desired result. So I
> need to know how to write the if statement to make it work. This should be
> simple, but I just don't know how to do it, never had this problem before.
> --
> http://mail.python.org/mailman/listinfo/python-list

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