如何在网址中对字符串进行正则表达式

Question

http://something.com/bOhxBeD,SyhyTGi,TMDDSIB,U72gx2J,kQTIRy9,7VXgGDw,eSxIcK6,S5oNlnn,WBHHsLk,BdMGd2d,U9kNlsF,cHVyc7Y,D83kaJ5,cLWgdSO,iWtCIF3,ount8L6

I have tried to get the value: bOhxBeD, SyhyTGi and so on. This is what I come up with ( yes fairly simple ) /([a-zA-Z0-9]{7})/ , it seems to work with PCRE:

([a-zA-Z0-9]{7})

Debuggex Demo

But when it comes to Ruby, I use it like this :

str.match(/([a-zA-Z0-9]{7})/)
#<MatchData "bOhxBeD" 1:"bOhxBeD">

it doesn't seem to work. Can anyone point out what's wrong with this regex ? Thanks

Answer 1

You need to add word boundary \\b inorder to match an exact 7 alphanumeric characters.

\b[a-zA-Z0-9]{7}\b

DEMO

irb(main):006:0> "http://something.com/bOhxBeD,SyhyTGi,TMDDSIB,U72gx2J,kQTIRy9,7VXgGDw,eSxIcK6,S5oNlnn,WBHHsLk,BdMGd2d,U9kNlsF,cHVyc7Y,D83kaJ5,cLWgdSO,iWtCIF3,ount8L6".scan(/\b([a-zA-Z0-9]{7})\b/)
=> [["bOhxBeD"], ["SyhyTGi"], ["TMDDSIB"], ["U72gx2J"], ["kQTIRy9"], ["7VXgGDw"], ["eSxIcK6"], ["S5oNlnn"], ["WBHHsLk"], ["BdMGd2d"], ["U9kNlsF"], ["cHVyc7Y"], ["D83kaJ5"], ["cLWgdSO"], ["iWtCIF3"], ["ount8L6"]]

Answer 2

 (?!.*?\/)[a-zA-Z0-9]{7}

应该是这个。否则它将从链接中也选择7个字母词。“ somethi”将在ans中。但是我想这不是必需的。

Answer 3

match only picks up the first match.
You can try the global version of match which is scan .
You can use scan to search string not containing specific characters using [^...] :

str.scan(/[^\/\.\,]+/)[3..-1]   
#=> ["bOhxBeD", "SyhyTGi", "TMDDSIB", "U72gx2J", "kQTIRy9", "7VXgGDw", "eSxIcK6", "S5oNlnn", "WBHHsLk", "BdMGd2d", "U9kNlsF", "cHVyc7Y", "D83kaJ5", "cLWgdSO", "iWtCIF3", "ount8L6"]  

Update:
If you know that the strings between the comma are always 7 characters, you can use this instead:

   str.scan(/[^\/\.\,]{7}/)[1..-1]

Answer 4

发生这种情况是因为您的正则表达式仅匹配一个包含7个字符的元素，仅此而已，因为简单的解决方案可能是：

str.match(/\/(.*)\z/)[1].split(',')

Answer 5

You could use String#[] and String#split :

str[/.*\/(.*)/,1].split(',')
  #=> ["bOhxBeD", "SyhyTGi", "TMDDSIB", "U72gx2J", "kQTIRy9", "7VXgGDw",
  #    "eSxIcK6", "S5oNlnn", "WBHHsLk", "BdMGd2d", "U9kNlsF", "cHVyc7Y",
  #    "D83kaJ5", "cLWgdSO", "iWtCIF3", "ount8L6"]

.*\\/ in the regex, "greedy" as it is, will consume characters up to and including the last forward slash in the string. Capture group #1 (.*) sucks up the remainder of the string and, due to the presence of ,1 , returns it. split(',') then breaks up the string to give you the desired array.

Another way:

str[str[/.*\//].size..-1].split(',')