I can't figure this out:
22.584\r\n\t\t\tl-6.579-22
I want to match the "\r\n\t\t\t" and replace with a single space " ". Problem is the number of "\t", "\r", and "\n" fluctuates, as do the surrounding characters.
Help!
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What about ordinary spaces? Do you also want to replace two spaces with one space?Mark Byers– Mark Byers2010-01-27 09:31:42 +00:00Commented Jan 27, 2010 at 9:31
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6 Answers
s/\s+/ /g
s/(?:\\[rnt])+/ /g
6 Comments
Bart Kiers
@neezer, note that
\s matches more than just \t, \r or \n. It matches (in most cases) these: [ \t\n\x0B\f\r], but that may be fine with you of course.
kennytm
@neezer: What language are you using?
Mark Byers
@neezer: In what way doesn't it work? Does it give an error? Does it replace too much? Too little? Does it do nothing at all?
neezer
I'm using Ruby. It doesn't match anything. The values in the string in my question are not escape values (as in, to match, you'd search for
\\r or the like).neezer
Your edit matches them individually, not the group. I need to replace the whole group with a single space. Any other thoughts?
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In PHP:
preg_replace("/(?:\\\[trn])+/", " ", $str);
2 Comments
glenn jackman
did you see this was a ruby question?
Matteo Riva
When I answered there was no mention of ruby, or any other language. The post was edited later, as you can see. Also, accepted answer uses Perl's syntax.
sed 's/\\[rnt]/ /g;s/ */ /g'
Comments
'22.584\r\n\t\t\tl-6.579-22'.gsub(/(\\[rnt])+/, ' ')
Comments
#!/usr/bin/ruby1.8
s = "22.584\r\n\t\t\tl-6.579-22"
p s # => "22.584\r\n\t\t\tl-6.579-22"
p s.gsub(/[\r\n\t]+/, ' ') # => "22.584 l-6.579-22"
Comments
I'd treat the CR-NL as one atom:
str.gsub!(/(?:\r\n)+\t+/, ' ')
