strip method

[Amasa Masiarek]

s = <<EOS
a1\t b1
a2\tb2
EOS

s.each{|e|
a = e.strip.split("\t")
p a[1]
}

output:
" b1"
"b2"

expected output:
"b1"
"b2"

What is wrong?
--
Posted via http://www.ruby-forum.com/.

[Amasa Masiarek]

This works fine ...

s.each{|e|
a = e.gsub(' ','').split("\t")
p a[1]
}

[Rick DeNatale]

2008/5/1 Amasa Masiarek <ad...@masiarek.com>:

> s = <<EOS
> a1\t b1
> a2\tb2
> EOS
>
> s.each{|e|
> a = e.strip.split("\t")
> p a[1]
> }
>
> output:
> " b1"
> "b2"
>
> expected output:
> "b1"
> "b2"
>
> What is wrong?

k$ qri String#strip
----------------------------------------------------------- String#strip
str.strip => new_str
------------------------------------------------------------------------
Returns a copy of str with leading and trailing whitespace removed.

Leading and trailing doesn't mean internal. So:

"a1\t b1".strip #=> "a1\t b1"

--
Rick DeNatale

My blog on Ruby
http://talklikeaduck.denhaven2.com/

[David A. Black]

Hi --

On Fri, 2 May 2008, Amasa Masiarek wrote:

> s = <<EOS
> a1\t b1
> a2\tb2
> EOS
>
> s.each{|e|
> a = e.strip.split("\t")
> p a[1]
> }
>
> output:
> " b1"
> "b2"
>
> expected output:
> "b1"
> "b2"
>
> What is wrong?

Your expectation :-) Sorry, I couldn't resist.

strip removes space on the right and left. If you start with
"a1\t b1\n", the stripped version is "a1\t b1". If you split that on
\t, you get "a1" and " b1". At no point have you removed the middle
space.

David

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[William James]

"

".strip.each{|x|
a = x.split("\t").map{|s| s.strip }
p a.last
}

[Gordon Thiesfeld]

Or split on whitespace instead of on tabs alone.

s.each{|e|
a = e.split("\s")
p a[1]
}