给定一个哈希值
PLATFORMS = {
:mac => /(mac)|(macintosh)/i,
:win => /(win)|(windows)/i,
:ipad => /(ipad)/i,
:iphone => /(iphone)/i,
:ipod => /(ipod)|(ipod touch)/i
}
我有兴趣返回散列键,其中值(在本例中是正则表达式)返回 true。
因此,如果我得到一个字符串 "windows"
,我应该返回 key :win
。
到目前为止我的尝试是:
current_platform = BrowserExperience::ExperienceKeeper::PLATFORMS.detect do |platform, regex|
regex.match(user_agent_obj.platform)
end[0]
它返回 [:win,/(win)|(windows)/i]
但是,它只返回一个数组,其中索引 0 返回我想要的键值。有没有更简单的方法?
最佳答案
为什么不使用 case
语句?这是一种更常见的方法:
strings = [
'this is a Windows box',
'Welcome to Macintosh',
'My music is on an iPod',
'My photos are on an iPod Touch',
'I read books on an iPad'
]
strings.each do |str|
os = case str
when /\b(?:mac|macintosh)\b/i
:mac
when /\b(?:win|windows)\b/i
:win
when /\b(?:ipad)\b/i
:ipad
when /\b(?:iphone)\b/i
:iphone
when /\b(?:ipod|ipod\ touch)\b/i
:ipod
end
os # => :win, :mac, :ipod, :ipod, :ipad
end
也可以这样做:
PLATFORMS = {
mac: /\b(?:mac|macintosh)\b/i,
win: /\b(?:win|windows)\b/i,
ipad: /\b(?:ipad)\b/i,
iphone: /\b(?:iphone)\b/i,
ipod: /\b(?:ipod|ipod\ touch)\b/i
}
strings.each do |str|
key = nil
PLATFORMS.each_pair do |k, v|
if str =~ v
key = k
break
end
end
key # => :win, :mac, :ipod, :ipod, :ipad
end
或者最好的:
strings.each do |str|
PLATFORMS.find { |k, v| str =~ v }.first # => :win, :mac, :ipod, :ipod, :ipad
end
如果您使用散列和正则表达式,请使用更简洁的模式。 \b
是一个单词边界,它是我们告诉 Regexp 引擎是匹配子串还是整个单词的方式:
'machine'[/(?:mac|macintosh)/i] # => "mac"
对比:
'machine'[/\b(?:mac|macintosh)\b/i] # => nil
还有一点:
'mac'[/\b(?:mac|macintosh)\b/i] # => "mac"
'macintosh'[/\b(?:mac|macintosh)\b/i] # => "macintosh"
'win'[/\b(?:win|windows)\b/i] # => "win"
'windows'[/\b(?:win|windows)\b/i] # => "windows"
'ipad'[/\b(?:ipad)\b/i] # => "ipad"
'iphone'[/\b(?:iphone)\b/i] # => "iphone"
'ipod touch'[/\b(?:ipod|ipod\ touch)\b/i] # => "ipod"
我可能会做这样的事情来定义散列:
require 'regexp_trie'
PLATFORMS = {
mac: ['mac', 'macintosh'],
win: ['win', 'windows'],
ipad: ['ipad'],
iphone: ['iphone'],
ipod: ['ipod', 'ipod touch']
}
然后,我会将模式转换为更高效的模式:
PLATFORMS_RE = {}
PLATFORMS.each_pair do |k, v|
PLATFORMS_RE[k] = /\b(?:#{RegexpTrie.union(v).source})\b/i
end
结果是:
PLATFORMS_RE
# => {:mac=>/\b(?:mac(?:intosh)?)\b/i,
# :win=>/\b(?:win(?:dows)?)\b/i,
# :ipad=>/\b(?:ipad)\b/i,
# :iphone=>/\b(?:iphone)\b/i,
# :ipod=>/\b(?:ipod(?:\ touch)?)\b/i}
然后像以前一样工作:
strings.each do |str|
PLATFORMS_RE.find { |k, v| str =~ v }.first # => :win, :mac, :ipod, :ipod, :ipad
end
关于ruby-on-rails - 如果值为真,如何返回哈希键?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/44465125/