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Extracting the last n characters from a ruby string

Answer 1

A:

ending = string.reverse[0...n].reverse

Andrew Grimm 2010-02-01 05:15:16

Answer 2

+11 A:

Well, the easiest workaround I can think of is:

ending = str[-n..-1] || str

(EDIT: The or operator has lower precedence than assignment, so be sure to use || instead.)

perimosocordiae 2010-02-01 05:17:36

+1... I think this way is easier to read than `string.reverse[0..n].reverse`, which gives me a second of "wait, why is he doing that?" (or would if I weren't reading it in the context of this question)

Arkaaito 2010-02-01 05:19:31

Much simpler than my regex.

EmFi 2010-02-01 05:27:11

Good answer, but it should be `||` instead of `or`, or put parentheses around `str[-n..-1] or str`.

Andrew Grimm 2010-02-01 05:34:24

Good answer, but I don't like that ruby doesn't treat s[-inf..-1] the same as x[0..inf]

klochner 2010-02-01 07:35:20

Thanks for noting the operator precedence issue, Andrew. Gets me every time.

perimosocordiae 2010-02-02 08:21:36

@perimosocordiae you aren't the only one. http://stackoverflow.com/questions/372652/what-are-the-ruby-gotchas-a-newbie-should-be-warned-about

Andrew Grimm 2010-02-02 22:06:53

Answer 3

+1 A:

Have you tried a regex?

string.match(/(.{0,#{n}}$)/)
ending=$1

The regex captures as many characters it can at the end of the string, but no more than n. And stores it in $1.

EmFi 2010-02-01 05:19:45

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Extracting the last n characters from a ruby string

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