why doesn't python return result using regex when there's more than one match? [duplicate]

Question

Here's the code:

pattern = re.compile(r'ea') 
match = pattern.match('sea ea')
if match:
    print match.group()

the result is null. But when I change the code to pattern = re.compile(r'sea'), the output is "sea"

Could anyone give me an explanation?

p.s. Btw, What I want is to retrieve the "#{year}" from string "select * from records where year = #{year}", plz give me an usable regex. Thanks in advance!

Summary:

Thanks to ALL of u, I find it in the document of python with your instruction. since I can select only one most appropriate answer, I just give it to the one who answered most quickly. Thx again.

devnull · Accepted Answer · 2013-10-10 05:44:29Z

2

pattern.match is anchored at the beginning of the string.

You need pattern.search.

From the documentation:

Python offers two different primitive operations based on regular expressions: re.match() checks for a match only at the beginning of the string, while re.search() checks for a match anywhere in the string (this is what Perl does by default).

answered Oct 10, 2013 at 5:44

devnull

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Comments

David Robinson · Accepted Answer · 2013-10-10 05:44:14Z

2

You mean to use search, not match. match will match the regular expression only if it is at the start of the string.

pattern = re.compile(r'ea') 
match = pattern.search('sea ea')
if match:
    print match.group()

answered Oct 10, 2013 at 5:44

David Robinson

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Comments

jhermann · Accepted Answer · 2013-10-10 05:47:42Z

0

match just matches, it doesn't search for things. This does:

>>> pattern = re.compile(r'(#{\w+})') 
>>> pattern.split('select * from records where year = #{year}')
['select * from records where year = ', '#{year}', '']

answered Oct 10, 2013 at 5:47

jhermann

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