Search or compai problem

[Gallagher, Tim (NE)]

I am new to python and I want to compare 2 strings, here is my code:
[start]

import active_director y
import re

lstUsers = []
users = active_director y.root()
for user in users.search ("sn='gallagher '"):
lstUsers.append (user.samAccoun tName)

print "----------------------------------------"
lstUsers.sort()

## Printing out what is inside of the arrar(list)
x = 0
while x < len(lstUsers):
if re.compile(lstU sers[x]).match("None", 0):
print "Somthing here"

x = x + 1

[/end]

When I do the:
if re.compile(lstU sers[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.
Thanks

Timothy F. Gallagher
CSC Systems Engineer

[John Machin]

Re: Search or compai problem

Gallagher, Tim (NE) wrote:

I am new to python and I want to compare 2 strings, here is my code:
[start]
>
import active_director y
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_director y.root()
for user in users.search ("sn='gallagher '"):

**** Insert here:
print type(user.samAc countName), repr(user.samAc countName)
**** that may indicate where your problem *starts*
**** Then read the documentation for the active_director y module, in
particular what it says about the attributes of the objects in the
sequence returned by users.search.

lstUsers.append (user.samAccoun tName)
>
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_wi th(item)

if re.compile(lstU sers[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(cons tant 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(lstU sers[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

[timgerr@gmail.com]

Re: Search or compai problem


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_director y
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_director y.root()
for user in users.search ("sn='gallagher '"):

>
**** Insert here:
print type(user.samAc countName), repr(user.samAc countName)
**** that may indicate where your problem *starts*
**** Then read the documentation for the active_director y module, in
particular what it says about the attributes of the objects in the
sequence returned by users.search.
>
>

lstUsers.append (user.samAccoun tName)

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_wi th(item)
>

if re.compile(lstU sers[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(cons tant 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(lstU sers[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

[Gabriel Genellina]

Re: Search or compai problem

At Saturday 19/8/2006 01:16, timgerr@gmail.c om wrote:

>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

None is a unique object used as "nothing" or "no value".
Try reading the Python tutorial, it's easy and you will learn a lot of things.


Gabriel Genellina
Softlab SRL





_______________ _______________ _______________ _____
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[timgerr@gmail.com]

Re: Search or compai problem


Gabriel Genellina wrote:

At Saturday 19/8/2006 01:16, timgerr@gmail.c om wrote:
>

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

>
None is a unique object used as "nothing" or "no value".
Try reading the Python tutorial, it's easy and you will learn a lot of things.
>
>
Gabriel Genellina
Softlab SRL
>
>
>
>
>

Thanks, I did not know that. Then I should look for a null value?

Tim

[Gabriel Genellina]

Re: Search or compai problem

At Saturday 19/8/2006 23:43, timgerr@gmail.c om wrote:

>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

None is a unique object used as "nothing" or "no value".
Try reading the Python tutorial, it's easy and you will learn a

lot of things.
>
>Thanks, I did not know that. Then I should look for a null value?

Yes; I don't know where your items come from, but usually None is
used to represent an empty/null value. It's not the same as "".


Gabriel Genellina
Softlab SRL





_______________ _______________ _______________ _____
Preguntá. Respondé. Descubrí.
Todo lo que querías saber, y lo que ni imaginabas,
está en Yahoo! Respuestas (Beta).
¡Probalo ya!

http://www.yahoo.com.ar/respuestas