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Regex Expression not matching correctly

I'm tackling a python challenge problem to find a block of text in the format xXXXxXXXx (lower vs upper case, not all X's) in a chunk like this:
jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn
I have tested the following RegEx and found it correctly matches what I am looking for from this site (http://www.regexr.com/):
'([a-z])([A-Z]){3}([a-z])([A-Z]){3}([a-z])'
However, when I try to match this expression to the block of text, it just returns the entire string:
In [1]: import re
In [2]: example = 'jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn'
In [3]: expression = re.compile(r'([a-z])([A-Z]){3}([a-z])([A-Z]){3}([a-z])')
In [4]: found = expression.search(example)
In [5]: print found.string
jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn
Any ideas? Is my expression incorrect? Also, if there is a simpler way to represent that expression, feel free to let me know. I'm fairly new to RegEx.

You need to return the match group instead of the string attribute.
>>> import re
>>> s = 'jdskvSJNDfbSJneSfnJDKoJIWhsjnfakjn'
>>> rgx = re.compile(r'[a-z][A-Z]{3}[a-z][A-Z]{3}[a-z]')
>>> found = rgx.search(s).group()
>>> print found
nJDKoJIWh

The string attribute always returns the string passed as input to the match. This is clearly documented:
string
The string passed to match() or search().
The problem has nothing to do with the matching, you're just grabbing the wrong thing from the match object. Use match.group(0) (or match.group()).

Based on xXXXxXXXx if you want upper letters with len 3 and lower with len 1 between them this is what you want :
([a-z])(([A-Z]){3}([a-z]))+
also you can get your search function with group()
print expression.search(example).group(0)

Python regex to match multiple times

I'm trying to match a pattern against strings that could have multiple instances of the pattern. I need every instance separately. re.findall() should do it but I don't know what I'm doing wrong.
pattern = re.compile('/review: (http://url.com/(\d+)\s?)+/', re.IGNORECASE)
match = pattern.findall('this is the message. review: http://url.com/123 http://url.com/456')
I need 'http://url.com/123', http://url.com/456 and the two numbers 123 & 456 to be different elements of the match list.
I have also tried '/review: ((http://url.com/(\d+)\s?)+)/' as the pattern, but no luck.

Use this. You need to place 'review' outside the capturing group to achieve the desired result.
pattern = re.compile(r'(?:review: )?(http://url.com/(\d+))\s?', re.IGNORECASE)
This gives output
>>> match = pattern.findall('this is the message. review: http://url.com/123 http://url.com/456')
>>> match
[('http://url.com/123', '123'), ('http://url.com/456', '456')]

You've got extra /'s in the regex. In python the pattern should just be a string. e.g. instead of this:
pattern = re.compile('/review: (http://url.com/(\d+)\s?)+/', re.IGNORECASE)
It should be:
pattern = re.compile('review: (http://url.com/(\d+)\s?)+', re.IGNORECASE)
Also typically in python you'd actually use a "raw" string like this:
pattern = re.compile(r'review: (http://url.com/(\d+)\s?)+', re.IGNORECASE)
The extra r on the front of the string saves you from having to do lots of backslash escaping etc.

Use a two-step approach: First get everything from "review:" to EOL, then tokenize that.
msg = 'this is the message. review: http://url.com/123 http://url.com/456'
review_pattern = re.compile('.*review: (.*)$')
urls = review_pattern.findall(msg)[0]
url_pattern = re.compile("(http://url.com/(\d+))")
url_pattern.findall(urls)

What should I use the Non-greedy match in this case

Assume I have a string which includes some data fields that are separated by "|", like
|1|2|3|4|5|6|7|8|
My purpose is to get the 8th field. This is what I'm doing:
pattern = re.compile(r'^\s+(\|.*?\|){8}')
match = pattern.match(test_line)
if match:
print:match.group(8)
But looks like it can not match. I know in this case I need to use ? for non-greedy match, but why I can not get the 8th field?
Thanks

Regex might be complicating this problem rather than simplifying it. A simple way to get an eighth item from a | delimited string is using split():
a = '|here|is|some|data|separated|by|bars|hooray!|'
print a.split('|')[8]
RETURNS
hooray!
Using regex, one way to get it would be:
import re
a = '|here|is|some|data|separated|by|bars|hooray!|'
pattern = re.compile(r'([^\|]+)')
match = pattern.findall(a)
print match[7]
RETURNS
hooray!

Simple python regex, match after colon

I have a simple regex question that's driving me crazy.
I have a variable x = "field1: XXXX field2: YYYY".
I want to retrieve YYYY (note that this is an example value).
My approach was as follows:
values = re.match('field2:\s(.*)', x)
print values.groups()
It's not matching anything. Can I get some help with this? Thanks!

Your regex is good
field2:\s(.*)
Try this code
match = re.search(r"field2:\s(.*)", subject)
if match:
result = match.group(1)
else:
result = ""

re.match() only matches at the start of the string. You want to use re.search() instead.
Also, you should use a verbatim string:
>>> values = re.search(r'field2:\s(.*)', x)
>>> print values.groups()
('YYYY',)

regex in python 2.4

I have a string in python as below:
"\\B1\\B1xxA1xxMdl1zzInoAEROzzMofIN"
I want to get the string as
"B1xxA1xxMdl1zzInoAEROzzMofIN"
I think this can be done using regex but could not achieve it yet. Please give me an idea.

st = "\B1\B1xxA1xxMdl1zzInoAEROzzMofIN"
s = re.sub(r"\\","",st)
idx = s.rindex("B1")
print s[idx:]
output = 'B1xxA1xxMdl1zzInoAEROzzMofIN'
OR
st = "\B1\B1xxA1xxMdl1zzInoAEROzzMofIN"
idx = st.rindex("\\")
print st[idx+1:]
output = 'B1xxA1xxMdl1zzInoAEROzzMofIN'

Here is a try:
import re
s = "\\B1\\B1xxA1xxMdl1zzInoAEROzzMofIN"
s = re.sub(r"\\[^\\]+\\","", s)
print s
Tested on http://py-ide-online.appspot.com (couldn't find a way to share though)
[EDIT] For some explanation, have a look at the Python regex documentation page and the first comment of this SO question:
How to remove symbols from a string with Python?
because using brackets [] can be tricky (IMHO)
In this case, [^\\] means anything but two backslashes \\.
So [^\\]+ means one or more character that matches anything but two backslashes \\.

If the desired section of the string is always on the RHS of a \ char then you could use:
string = "\\B1\\B1xxA1xxMdl1zzInoAEROzzMofIN"
string.rpartition("\\")[2]
output = 'B1xxA1xxMdl1zzInoAEROzzMofIN'