[ruby-list:50346] Re: usage of `?<='

5.5 です。

a.split("@foo") ではだめでしょうか。（これだと先頭に空文字列
を拾っちゃいますが）
意図を誤解してたらすみません。
もう少し詳細な仕様がわかるといいのですが。

On 2016/07/07 7:18, KIRIYAMA Kazuhiko wrote:
> すいません．Subject タイポしてました．
>
> At Thu, 07 Jul 2016 07:11:05 +0900,
> 私 wrote:
>>
>> しばらく Ruby を使ってなくて，ちょっと解決できない問題があり
>> ました．肯定的先読み演算子 `？<=' の使い方です．
>
> ↑は「肯定的先読み」→「肯定的後読み」の間違いです．
>
>>
>> kiri@kazu:~[1007]% ruby --version
>> ruby 1.9.3p484 (2013-11-22 revision 43786) [amd64-freebsd9]
>> kiri@kazu:~[1008]% irb
>> irb(main):001:0> a = '@foo
>> irb(main):002:0' This is 1st foo line 1
>> irb(main):003:0' This is 1st foo line 2
>> irb(main):004:0' This is 1st foo line 3
>> irb(main):005:0' @foo
>> irb(main):006:0' This is 2nd foo line 1
>> irb(main):007:0' @foo
>> irb(main):008:0' This is 3rd foo line 1
>> irb(main):009:0' This is 3rd foo line 2
>> irb(main):010:0' '
>> => "@foo\nThis is 1st foo line 1\nThis is 1st foo line 2\nThis is 1st foo line 3\n@foo\nThis is 2nd foo line 1\n@foo\nThis is 3rd foo line 1\nThis is 3rd foo line 2\n"
>> irb(main):011:0> a.scan(/(@foo(?:.|\n)*?)(?<=(@foo|\z))/).each do |content|
>> irb(main):012:1* p content
>> irb(main):013:1> end
>> SyntaxError: (irb):11: invalid pattern in look-behind: /(@foo(?:.|\n)*?)(?<=(@foo|\z))/
>>         from /usr/local/bin/irb:12:in `<main>'
>> irb(main):014:0> a.scan(/(@foo(?:.|\n)*?)(?<=@foo|\z)/).each do |content|
>> irb(main):015:1* p content
>> irb(main):016:1> end
>> SyntaxError: (irb):14: invalid pattern in look-behind: /(@foo(?:.|\n)*?)(?<=@foo|\z)/
>>         from /usr/local/bin/irb:12:in `<main>'
>> irb(main):017:0> a.scan(/(@foo(?:.|\n)*?)(?<=@foo)/).each do |content|
>> irb(main):018:1* p content
>> irb(main):019:1> end
>> ["@foo"]
>> ["@foo"]
>> ["@foo"]
>> => [["@foo"], ["@foo"], ["@foo"]]
>> irb(main):020:0>
>>
>> となって，どうもうまく機能しないのですが，a の @foo から次の
>> @foo までの文字列を順番に取り出すにはどうすれば良いですか？
>>
>> ---
>> KIRIYAMA Kazuhiko
>>
>


-- 
[e-mail]