I need to remove a string A that exists in another string B only if string A is between two spaces.
string A = "e"
string B = "the fifth letter is e "
Example for replacing 'e' : "the fifth letter is e " --> "the fifth letter is"
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I don't get your example. Which one is A and B here? Are there restrictions? like the string being surrounded by more than one space? In your example e is not between two spaces. What is the replacement? removing it?duncan– duncan2011-10-10 19:25:16 +00:00Commented Oct 10, 2011 at 19:25
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3 Answers
ruby-1.9.2-p290 :006 > a = "the fifth letter is e "
=> "the fifth letter is e "
ruby-1.9.2-p290 :007 > print a.gsub(/\se\s/,"")
the fifth letter is => nil
Edited the answer after you edited the question. A possible regular expression to find an "e" character between two space characters is /\se\s/. In this case I'm replacing it with an empty string "". You can use gsub which returns a copy of the string or gsub! to modify the original string.
UPDATE: Since you edited the question again, here's un updated answer:
ruby-1.9.2-p290 :001 > a = "e"
=> "e"
ruby-1.9.2-p290 :002 > b = "the fifth letter is e "
=> "the fifth letter is e "
ruby-1.9.2-p290 :003 > print b.gsub(/\s#{a}\s/,"")
the fifth letter is => nil
1 Comment
Fernando Briano
Yeah, it is wrong because the question is constantly being changed, so I should be constantly updating the answer...
You don't really need regex for this.
a = "e"
b = "the fifth letter is e "
c = b.gsub(" " << a << " ", "")
PS. In Ruby it's a constant if it begins with an uppercase letter. DS.
Comments
str = 'the fifth letter is e'
thing = 'e'
str.sub! /\s+#{thing}\s+/, ''
1 Comment
C. K. Young
It works for me. Of course, I've edited the post a few times in between.... (Standard FGitW technique, y'know? ;-))