Search or compai problem

[timgerr at gmail.com]

John Machin wrote:
> Gallagher, Tim (NE) wrote:
> > I am new to python and I want to compare 2 strings, here is my code:
> > [start]
> >
> > import active_directory
> > import re
> >
> > lstUsers = []
>
> Using "lst" or "l" as a variable name is bad news in any language; with
> many fonts they are too easily misread as "1st" or "1" respectively.
>
> > users = active_directory.root()
> > for user in users.search ("sn='gallagher'"):
>
> **** Insert here:
>         print type(user.samAccountName), repr(user.samAccountName)
> **** that may indicate where your problem *starts*
> **** Then read the documentation for the active_directory module, in
> particular what it says about the attributes of the objects in the
> sequence returned by users.search.
>
>
> >     lstUsers.append(user.samAccountName)
> >
> > print "----------------------------------------"
> > lstUsers.sort()
> >
> > ## Printing out what is inside of the arrar(list)
>
> What is "arrar(list)" ??
>
> **** Here insert this code:
> print lstUsers
> **** Look at the elements in the list; do you see ..., 'None', ... or
> do you see ..., None, ...
>
> > x = 0
> > while x < len(lstUsers):
>
> *Please* consider using a "for" statement:
>
> for item in lstUsers:
>     do_something_with(item)
>
> >     if re.compile(lstUsers[x]).match("None",0):
>
> 1. Python or not, using regular expressions to test for equality is
> extreme overkill. Use
>     if lstUsers[x] == "None":
> 2. Python or not, it is usual to do
>     re.compile(constant pattern).match(variable_input)
> not the other way around.
> 3. The second arg to match defaults to 0, so you can leave it out.
>
>
> >         print "Somthing here"
> >
> >     x = x + 1
> >
> > [/end]
> >
> > When I do the:
> >     if re.compile(lstUsers[x]).match("None",0):
> >         print "Somthing here"
> >
> > Some of the items in lstUsers[x] are the word None.
> >  I am not sure why I cant do this
> >
> > I want to compare lstUsers[x] to the word "None", how can I do this.
>
> You  *have* compared lstUsers[x] to the word "None" -- with the re
> sledgehammer, but you've done it. Maybe what's in there is *not* the
> string "None" :-)
>
> HTH,
> John

it is really lstusers (it is an L not a # 1),   Some of the output from
print lstUsers has the output of None.  I and trying to filter the None
out of the list.  I come from a perl background and this is how I do
thing in perl

TIm