case1 = http:www.freejobalert.comupsc-advt-no-18337
case2 = http:www.freejobalert.comupsc-advt-no-1833742
job_id = (''.join(re.findall(r'\d{7}:\d{5}',case1,re.I)))
how can i find only 33742 from this above string using regular expression . the number maybe 5 or 7 character.
The issue is that you are not using "or" (|), but instead a colon; Try using:
\d{7}|\d{5}
It should return the number in your string.
You can also try something like \d{5,7} meaning between 5 and 7 (inclusive) digits.
job_id = (''.join(re.search(r'(\d{5,7})',case1,re.I)))
The default behavior will be greedy, meaning it will match as many digits as possible between 5 and 7 digits.
EDIT (Another option that may be easier to understand but will do the same thing):
job_id = re.search(r'(\d{5,7})',case1).group(1)
Both the before mentioned answers are solutions to your problem, i have sumrises then as follows:
>>> import re
>>> case1 = "http:www.freejobalert.comupsc-advt-no-18337"
>>> case2 = "http:www.freejobalert.comupsc-advt-no-1833742"
>>> job_id1 = (''.join(re.findall(r'\d{5,7}',case1,re.I)))
>>> job_id2 = (''.join(re.findall(r'\d{5,7}',case2,re.I)))
>>> job_id1
'18337'
>>> job_id2
'1833742'
>>> job_id3 = (''.join(re.findall(r'\d{5}|\d{7}',case1,re.I)))
>>> job_id4 = (''.join(re.findall(r'\d{5}|\d{7}',case2,re.I)))
>>> job_id3
'18337'
>>> job_id4
'18337'
\d{7}|\d{5}...