如果你想坚持字符串操作,你可以使用正则表达式和gsub。
这是一种方法。它可以使用一些清理(例如错误处理,重新分解等),但我认为它会让你开始。
def count(details, location, age_group)
location_details = /#{location}(.+?);/.match(details)[1]
age_count = /#{age_group}#(\d+)\{/.match(details)[1]
return age_count.to_i
end
def ids(details, location, age_group)
location_details = /#{location}(.+?);/.match(details)[1]
age_ids = /#{age_group}#\d+\{(.+?)\}/.match(details)[1]
return age_ids
end
def add(details, location, age_group, new_id)
location_details = /#{location}(.+?);/.match(details)[1]
new_count = count(details, location, age_group) + 1
new_ids = ids(details, location, age_group) + ',' + new_id
location_details.gsub!(/#{age_group}#\d+\{(.+?)\}/, "#{age_group}##{new_count}{#{new_ids}}")
details.gsub!(/#{location}(.+?);/, "#{location}#{location_details};")
end
你可以看到它产生你想要的结果(至少在功能上,不知道性能):
names = {"john doe" => "ca:tw#2{1,2}:th#1{3};ar:tw#1{4}:fi#1{5};ny:tw#1{6};"}
puts count(names["john doe"], 'ca', 'tw')
#=> 2
puts ids(names["john doe"], 'ca', 'tw')
#=> 1,2
names["john doe"] = add(names["john doe"], 'ca', 'tw', '100')
puts names["john doe"]
#=> ca:tw#3{1,2,100}:th#1{3};ar:tw#1{4}:fi#1{5};ny:tw#1{6};
你为什么要使用一个字符串,而不是存储更适合数据结构的这个数据? – 2012-08-14 20:47:59
根据您在做什么,它可能有助于导入数据并将其转换为XML文档。像Nokogiri这样的宝石可以很快地与它一起工作。 – 2012-08-15 14:21:01