红宝石字符串分割问题

Question

我有这个字符串： “asdasda = asdaskdmasd & asmda = asdasmda & ACK =成功& asdmas = asdakmsd & asmda = adasda”

我想要的ACK之间后得到的值&象征，ACK和&符号之间的值可以改变...

感谢

我想在红宝石的解决方案。

Answer 1

require "cgi" 

query_string = "asdasda=asdaskdmasd&asmda=asdasmda&ACK=Success&asmda=asdakmsd" 
parsed_query_string = CGI.parse(query_string) 
#=> { "asdasda" => ["asdaskdmasd"], 
#  "asmda" => ["asdasmda", "asdakmsd"], 
#  "ACK" => ["Success"] } 

parsed_query_string["ACK"].first 
#=> "Success" 

如果你也想（用URL的其余部分特别合）重建查询字符串，我会建议寻找到了addressable宝石。

require "addressable/uri" 

# Note the leading '?' 
query_string = "?asdasda=asdaskdmasd&asmda=asdasmda&ACK=Success&asmda=asdakmsd" 
parsed_uri = Addressable::URI.parse(query_string) 
parsed_uri.query_values["ACK"] 
#=> "Success" 

parsed_uri.query_values = parsed_uri.query_values.merge("ACK" => "Changed") 
parsed_uri.to_s 
#=> "?ACK=Changed&asdasda=asdaskdmasd&asmda=asdakmsd" 
# Note how the order has changed and the duplicate key has been removed due to 
# Addressable's built-in normalisation. 

Answer 2

s = "asdasda=asdaskdmasd&asmda=asdasmda&ACK=Success&asdmas=asdakmsd&asmda=adasda" 
m = s.match /.*ACK=(.*?)&/ 
puts m[1] 

，只是为了好玩无正则表达式：

Hash[s.split("&").map{|p| p.split("=")}]["ACK"] 

Answer 3

"asdasda=asdaskdmasd&asmda=asdasmda&ACK=Success&asdmas=asdakmsd&asmda=adasda"[/ACK=([^&]*)&/] 
$1 # => 'Success' 

Answer 4

一个快速的方法：

s = "asdasda=asdaskdmasd&asmda=asdasmda&ACK=Success&asdmas=asdakmsd&asmda=adasda" 
s.gsub(/ACK[=\w]+&/,"ACK[changedValue]&") 
    #=> asdasda=asdaskdmasd&asmda=asdasmda&ACK[changedValue]&asdmas=asdakmsd&asmda=adasda