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<  Erlang  ~  regexp with \.
5hundy
Posted: Tue Oct 23, 2007 8:39 am Reply with quote
User Joined: 14 Oct 2007 Posts: 17
I want the first part of a string before the "."

For example I want "foo" from "foo.abc"

in ruby I would do:
Code:
"foo.abc".gsub(/\..*/,'')


in erlang I tried:
Code:
regexp:sub("foo.abc", "\..*", "").


but I get an empty list. In many other cases it seems that \. does not work properly. Is this a bug? How should I split my string?
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hao
Posted: Tue Oct 23, 2007 1:10 pm Reply with quote
User Joined: 20 Aug 2007 Posts: 18 Location: Uppsala, Sweden
Hi, 5hundy,

Don't worry, '\.' is not a bug. It seems that you misused it. Please refer to: http://www.erlang.org//doc/man/regexp.html
At the end of this document about regexp module, you can find out some regular expression rules. Two of them need to be pay attention to:

Quote:
\c matches the escape sequence or literal character c.

. matches any character.


So '\.' could not be able to represent '.' in the string. I think the code should be modified to:

Code:
43> regexp:sub("foo.abc", "[.].*", "").
{ok,"foo",1}

44> regexp:sub("foo.abc123!@#$%^&*()_+=<>|", "[.].*", "").
{ok,"foo",1}


If you are not compulsory to use regexp module to achieve it, you can also choose string module. Code can be looked like the followings:

Code:
49> string:left("foo.abc",(string:chr("foo.abc", $.)-1)).
"foo"


or

Code:
51> string:tokens("foo.abc", ".").
["foo","abc"]
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5hundy
Posted: Tue Oct 23, 2007 7:39 pm Reply with quote
User Joined: 14 Oct 2007 Posts: 17
Got it. Yeah, I saw those lines in the man page but I had trouble reconciling that with what I know of regular expressions in other environments.

Thanks for the help! Very Happy
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hao
Posted: Tue Oct 23, 2007 9:45 pm Reply with quote
User Joined: 20 Aug 2007 Posts: 18 Location: Uppsala, Sweden
With pleasure. Glad to be helpful. Smile
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