lua中的正则表达式模式问题

发布时间：2020-12-14 21:42:42 所属栏目：大数据来源：网络整理

导读：我有一个URL字符串,需要从匹配中获取某些单词. 例： /school/student/studentname1/detail/55/address/address1 我能够传递获取所需的细节,如, local s1,s2,s3 =myString:match("/school/student/(.-)/detail/(.-)/address/(.-)") 现在的问题是我的字符串可

我有一个URL字符串,需要从匹配中获取某些单词.

例：

/school/student/studentname1/detail/55/address/address1

我能够传递获取所需的细节,如,

local s1,s2,s3 =myString:match("/school/student/(.-)/detail/(.-)/address/(.-)")

现在的问题是我的字符串可以

myString = /school/student/studentname1

要么

myString = /school/student/studentname1/detail/55

在那种情况下,我的正则表达式失败,任何帮助？

解决方法

local all_fields = { student = 1,detail = 2,address = 3 }

local function parse(str)
   local info = {}
   local index
   for w in str:gmatch"/([^/]+)" do
      if index then
         info[index] = w
         index = nil
      else
         index = all_fields[w]
      end
   end
   return (table.unpack or unpack)(info,1,3)
end

local myString = '/school/student/studentname1/detail/55/address/address1'
local s1,s3 = parse(myString)
print(s1,s3)

myString = '/school/student/studentname1/address/address1'
s1,s3)

myString = '/school/student/studentname1/detail/55'
s1,s3)

（编辑：李大同）

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