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我有兴趣从命令行查询 Ruby的最新稳定版本。我主要对"1.9.3p327" 形式的字符串感兴趣。

更新#1

澄清一下，目标是始终查询最新的稳定版本，无论是 1.9.3p327 还是 3.0.2p392。

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4 回答 4

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就像是：

curl 'http://ftp.ruby-lang.org/pub/ruby/1.9/' | ruby ./extract-and-print-max-patchlevel.rb

脚本的实现extract-and-print-max-patchlevel.rb是读者的练习，但这是一个开始：

#!/usr/bin/env ruby
maxpatch=0
maxstr=nil
STDIN.each_line do |line|
  next unless line =~ /1\.9\.3-p(\d+)\b/
  patch = $1.to_i
  if patch > maxpatch
    maxpatch = patch
    maxstr = $&
  end
end
puts maxstr

请注意，它假定 Ruby 1.9.3 是最新的，因此您可能需要重新访问它。

于 2013-02-04T17:46:08.123 回答
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Came back to this after starting to use rbenv. If you have rbenv installed, the following one-liner does the trick:

#!/usr/bin/env ruby

puts `~/.rbenv/bin/rbenv install --list`.split("\n").map{|item| item.strip}.select{|item| item[/^\d*\.\d*\.\d*/]}.reject{|item| (item.include? '-') && !(item =~ /-p\d*$/)}.last
于 2014-03-07T01:32:24.730 回答
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你可以玩sed或awk。例如：

curl "http://ftp.ruby-lang.org/pub/ruby/1.9/" | sed -E 's/^.*"(ruby-)(1.9.3-p[0-9]+)(.*)".*$/\2/' | sed -e '/^1.*/!d' | sort | sed '$!d'

我不是sed专家，这里可能是更好的解决方案，但这条线有效。

于 2013-02-04T18:37:21.210 回答
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在 @maerics' 和 @YevgeniyAnfilofyev 的答案之间进行选择，我发现 Ruby 更真实，并将其变成单行，假设 curl 中的列表已经排序，所以我们只需要最后一行满足条件：

curl 'http://ftp.ruby-lang.org/pub/ruby/1.9/' 2> /dev/null | ruby -e "puts STDIN.lines.map { |x| /1\.9\.3-p\d+\b/.match(x) }.compact.last[0]"

或者，Ruby 部分扩展：

puts STDIN.lines.map do |x|
  /1\.9\.3-p\d+\b/.match(x) # After such map we will get an array of MatchData or nils
end.compact.last[0] # Then we remove nils using Array#compact,
                    # get last MatchData and retrieve matched string with [0]

遵循一个良好的传统，给这个脚本添加排序作为练习留给读者:)

UPD: Of course, hard-coded 1.9.3 is bad, but if we replace each digit 1, 9 and 3 with \d, the regex seems to become more or less independent. The other problem is that we only look into .../ruby/1.9 folder of that ftp. This may be fixed if we, instead, look into .../ruby folder first, and find all version-numbered folders with regex /\d\.\d\//. Then repeat the above query, joining results from all folders. But, of course, this already cannot made into a one-liner...

于 2013-02-04T18:57:22.233 回答