Parentheses capture - anything that matches a parenthesized portion of
a regular expression is returned as part of the result of the match:
user=> (re-seq #"a(.)c" "abc")
(["abc" "b"])
If you don't want that behavior, you can use the special non-capturing
syntax, (?:...):
user=> (re-seq #"a(?:.)c" "abc")
("abc")
You don't have to escape pipes or any other special characters inside
a character class (that is, between [...]), because characters lose
their special meanings there: [.*] matches either a period or an
asterisk and has no relationship to the "any character" symbol or
"zero or more" repetition operator.
The only special things inside a character class are a leading '^',
which negates the class, and a '-' in the middle, which makes a range:
[^a-z] matches any single character that is not a lowercase letter (of
the English alphabet). Position matters: [-^] matches a literal
hyphen or caret, and [] is not an empty character class but a syntax
error (an unclosed character class that so far includes a literal ']'
character).
On Tue, Mar 30, 2010 at 12:37 PM, Glen Rubin <rubing...@gmail.com> wrote:
> The result is a little bit strange still, since I am getting
> dupliates. First, it returns the string I want
>
> 49|00|12 .... 12|a9|a4|ff
>
> but then it also returns the same string without the first and last 4
> characters, e.g.
>
> 12|....12|a9|
>
> Also, how come I don't need to escape the | inside the parenthesis?
>
> thanks Meikel!!
>
>
> On Mar 30, 10:59 am, Meikel Brandmeyer <m...@kotka.de> wrote:
>> Hi,
>>
>> you have to escape the |.
>>
>> user=> (re-seq #"49\|00\|([0-9a-f|]+)\|a4\|ff" "a5|a5|49|23|49|00|12|
>> fc|5e|a4|ff|a7|49|00|ee|d3|a4|ff|ae")
>> (["49|00|12|fc|5e|a4|ff|a7|49|00|ee|d3|a4|ff" "12|fc|5e|a4|ff|a7|49|00|
>> ee|d3"])
>>
>> However this will be greedy...
>>
>> Sincerely
>> Meikel
>
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