Possessive generic quantifier {m,n}+ not implemented in Ruby 1.9.3?

Possessive quantifiers are greedy and refuse backtrack. A regex /.{1,3}+b/ should mean: Match any character except line breaks, 1 to 3 times, as many as possible and don't backtrack. Tthen match the character b . In this example: 'ab'.sub /.{1,3}+b/, 'c' #=> "c" no substitution should take place, contrary to fact. The result in these two examples differs: 'aab'.sub /.{0,1}+b/,

在Ruby 1.9.3中没有实现的所有通用量词{m,n} +?

拥有量词是贪婪的,拒绝回溯。 正则表达式/.{1,3}+b/应该表示:尽可能多地匹配除换行符之外的任何字符,1到3次,并且不要回溯。 那么匹配字符b 。 在这个例子中: 'ab'.sub /.{1,3}+b/, 'c' #=> "c" 不应该发生替代,与事实相反。 这两个例子的结果不同: 'aab'.sub /.{0,1}+b/, 'c' #=> "c" 'aab'.sub /.?+b/, 'c' #=> "ac" 与Scala进行比较,他们给出了相同的答案: scala> ".{0,1}+b".r.rep

What does this regex `str.gsub(/\#{(.*?)}/)` do?

This question already has an answer here: Greedy vs. Reluctant vs. Possessive Quantifiers 7 answers .* is a greedy match, whereas .*? is a non-greedy match. See this link for a quick tutorial on them. Greedy matches will match as much as they can, while non-greedy matches will match as little as they can. In this example, the greedy variant grabs everything between the first { and the la

这个正则表达式`str.gsub(/\#{((*?)}/)`做了什么?

这个问题在这里已经有了答案: 贪婪与不愿意与拥有量词7答案 .*是一个贪婪的匹配,而.*? 是一个非贪婪的比赛。 看到这个链接的快速教程。 贪婪的比赛会尽可能地匹配,而非贪婪的比赛会尽可能少地匹配。 在这个例子中,贪婪的变体抓住了第一个{和最后一个} (最后一个右括号)之间的所有内容: 'start #{this is a match}{and so is this} end'.match(/#{(.*)}/)[1] # => "this is a match}{and so is this" 而非贪

Match comma separated list with Ruby Regex

Given the following string, I'd like to match the elements of the list and parts of the rest after the colon: foo,bar,baz:something Ie I am expecting the first three match groups to be "foo", "bar", "baz". No commas and no colon. The minimum number of elements is 1, and there can be arbitrarily many. Assume no whitespace and lower case. I've tried th

使用Ruby Regex匹配逗号分隔的列表

鉴于以下字符串,我想匹配列表中的元素和其余部分在冒号后面: FOO,酒吧,巴兹:东西 也就是说,我期待前三个比赛组是“foo”,“bar”,“baz”。 没有逗号,也没有冒号。 元素的最小数量是1,并且可以是任意多的。 假设没有空白和小写。 我试过这个,应该可以工作,但不会因为某种原因填充所有的比赛组: ^([az]+)(?:,([az]+))*:(something) 这与 1中的foo和 2中的baz(或任何最后一个元素)匹配。 我不明白为什么我

homebrew not working on OSX

Getting this error when i run the brew command on terminal. Need help resolving. /System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/rubygems/core_ext/kernel_require.rb:55:in `require': cannot load such file -- utils/popen (LoadError) from /System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/rubygems/core_ext/kernel_require.rb:55:in `require' from /us

自制软件不能在OSX上工作

在终端上运行brew命令时出现此错误。 需要帮助解决。 /System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/rubygems/core_ext/kernel_require.rb:55:in `require': cannot load such file -- utils/popen (LoadError) from /System/Library/Frameworks/Ruby.framework/Versions/2.0/usr/lib/ruby/2.0.0/rubygems/core_ext/kernel_require.rb:55:in `require' from /usr/local/Library/Homebrew/utils.

Regex with named capture groups getting all matches in Ruby

I have a string: s="123--abc,123--abc,123--abc" I tried using Ruby 1.9's new feature "named groups" to fetch all named group info: /(?<number>d*)--(?<chars>s*)/ Is there an API like Python's findall which returns a matchdata collection? In this case I need to return two matches, because 123 and abc repeat twice. Each match data contains of detail of each named

带有命名捕获组的正则表达式在Ruby中获得所有匹配

我有一个字符串: s="123--abc,123--abc,123--abc" 我尝试使用Ruby 1.9的新功能“命名组”来获取所有命名的组信息: /(?<number>d*)--(?<chars>s*)/ 是否有像Python的findall这样的API返回matchdata集合? 在这种情况下,我需要返回两个匹配,因为123和abc重复两次。 每个匹配数据包含每个命名捕获信息的详细信息,因此我可以使用m['number']来获取匹配值。 命名捕捉仅适用于一个匹配结果。 Ruby的f

Using $1, $2, etc. global variables inside method definition

Given the following two pieces of code: def hello(z) "hello".gsub(/(o)/, &z) end z = proc {|m| p $1} hello(z) # prints: nil def hello z = proc {|m| p $1} "hello".gsub(/(o)/, &z) end hello # prints: "o" Why are the outputs of these two pieces of code different? Is there a way to pass a block to gsub from outside of the method definition so that the variables $1 , $2 would be evalu

在方法定义中使用$ 1,$ 2等全局变量

鉴于以下两段代码: def hello(z) "hello".gsub(/(o)/, &z) end z = proc {|m| p $1} hello(z) # prints: nil def hello z = proc {|m| p $1} "hello".gsub(/(o)/, &z) end hello # prints: "o" 为什么这两段代码的输出不同? 有没有办法从方法定义之外将块传递给gsub ,这样变量$1 , $2将以与块在方法定义内部一样的方式进行计算? 为什么输出不同? Ruby中的proc具有词汇范围。 这意味着当它找到一个未

Skipping native extension recompilation on subsequent bundle install

My production deployments take a few extra minutes due to time it takes to install nokogiri gem (1.6.0). I understand this is because installing the gem triggers native extension compilation. Note that I have packaged my bundle and checked it into DVCS bundle package Is there a way to avoid recompilation of native extensions if nothing else has changed, so that deployments are faster? Upda

在后续的bundle安装上跳过本机扩展重新编译

由于安装nokogiri gem(1.6.0)需要时间,因此我的生产部署需要几分钟的时间。 我明白这是因为安装gem触发本地扩展编译。 请注意,我打包了我的包并将其检入DVCS bundle package 如果没有其他更改,是否有避免重新编译本机扩展的方法,以便部署更快? 更新: 我使用Opscode Chef部署(主厨独奏具体) 环境是:Ubuntu 12.04LTS 64bit Ruby 193-p448 我找到了一种方法来做到这一点。 这里是解释: Bundler默认将ge

Limit cucumber steps or steps file to specified feature or tag

I've written a modal slideshow for our web app that presents a navigation for sets of documents and exposes various metadata for those documents. This is a large component of the application with esoteric requirements, so I think it's fair enough that its core scenarios (given to me as acceptance criteria) should be both numerous yet internally consistent. To avoid having a new step f

将黄瓜步骤或步骤文件限制为指定的功能或标记

我为我们的Web应用程序编写了一个模式幻灯片,其中显示了一组文档的导航并为这些文档提供了各种元数据。 这是具有深奥要求的应用程序中的一个重要组成部分,所以我认为它的核心场景(作为接受标准给出的)应该既众多又内部一致,这很公平。 为了避免为我们的许多场景采取新的步骤,我已经调整了一个帮助器,它将人类可读的术语(如document caption转换为选择器: module SelectorsHelper def selector_for(term) case

How do I use the conditional operator (? :) in Ruby?

How is the conditional operator ( ? : ) used in Ruby? For example, is this correct? <% question = question.size > 20 ? question.question.slice(0, 20)+"..." : question.question %> It is the ternary operator, and it works like in C (the parenthesis are not required). It's an expression that works like: if_this_is_a_true_value ? then_the_result_is_this : else_it_is_this However,

如何在Ruby中使用条件运算符(?:)?

Ruby中使用条件运算符( ? : :)的方式如何? 例如,这是正确的吗? <% question = question.size > 20 ? question.question.slice(0, 20)+"..." : question.question %> 它是三元运算符,它和C一样工作(不需要括号)。 这是一个表达式,其工作原理如下: if_this_is_a_true_value ? then_the_result_is_this : else_it_is_this 但是,在Ruby中, if也是这样的表达式: if a then b else c end === a ? b : c

opened nested module anomaly in Ruby

Why does re-opening a nested module give different results depending on the syntax used? For example, this works fine: module A module E end end module A module E def E.e end end end But this: module A module E end end module A::E def E.e end end gives the error: reopen.rb:6:in `<module:E>': uninitialized constant A::E::E (NameError) from reopen.rb:5:in `<ma

在Ruby中打开嵌套的模块异常

为什么根据使用的语法重新打开嵌套模块会得到不同的结果? 例如,这工作正常: module A module E end end module A module E def E.e end end end 但是这个: module A module E end end module A::E def E.e end end 给出错误: reopen.rb:6:in `<module:E>': uninitialized constant A::E::E (NameError) from reopen.rb:5:in `<main>' (在有人指出这点之前,解决方法是在定义Ee