从表达式树中解析和提取属性所需的正则表达式帮助

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问题描述

这是一个有效的属性树表达式(它可以是递归的):

Here is a valid property tree expression (it can be recursive):

rootProperty:(prop1, prop2, subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3), prop3, etc)

因此实际上一个属性可以有许多属性和子属性.从这个表达式中，我想捕获以下内容:

So in effect a property can have many properties and sub-properties.From this expression I would like to capture the following:

根属性
prop1
prop2
subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3)
prop3

我尝试了几种方法，但无法以递归方式进行重复.因此寻求帮助.

I tried few approaches but could't get the repetitions working recursively. Hence seeking help.

谢谢坎南

推荐答案

由于递归(平衡括号)，这不是正则语言，因此正则表达式可能不是您所需要的.但假设您知道自己在做什么:

This is not a regular language due to recursion (balanced parens), so a regular expression might not be what you need. But assuming you know what you are doing:

([^:(), ]+)(?::\(((?R)?(?:, ?(?R))*)\))?

首先我们捕获属性的名称:一个或多个不是 :(), 的字符.

First we capture the name of the property: one or more characters that are not :(), .

([^:(), ]+)

一个属性可能有也可能没有子树，所以下一部分是可选的子树:

A property may or may not have a subtree, so the next part is the optional subtree:

(?:           <--- do not capture
   :          <--- literal ':'
   \(         <--- literal '('
      ...     <--- some stuff inside
   \)         <--- literal ')'
)?            <--- it is optional

里面的东西捕获了一个属性列表:

The stuff inside captures a list of properties:

(             <--- do capture
 (?R)         <--- recursively match a property
 (?:          <--- do not capture
    , ?       <--- comma followed by optional space
    (?R)      <--- recursively match another property
 )*           <--- any number of comma separated properties
)             <--- end capture

对于您的示例输入:

Input:
    rootProperty:(prop1, prop2, subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3), prop3, etc)
Match 1:
    rootProperty:(prop1, prop2, subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3), prop3, etc)
    Group 1:
        rootProperty
    Group 2:
        prop1, prop2, subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3), prop3, etc

然后您可以递归匹配每个匹配项的第二组以捕获子树的属性.应该有一种方法可以获取回溯信息，这样您就不需要这样做，但我不知道如何.

You could then recursively match the second group of each match for capturing the properties of a subtree. There should be a way to get the backtracking information so you don't need to do this, but I don't know how.

Input:
    prop1, prop2, subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3), prop3, etc
Match 1:
    prop1
Match 2:
    prop2
Match 3:
    subProp1:(prop1,subSubProp1:(prop1,prop2,etc),prop3)
    Group 1:
        subProp1
    Group 2:
        prop1,subSubProp1:(prop1,prop2,etc),prop3
Match 4:
    prop3
Match 5:
    etc

那么，

Input:
    prop1,subSubProp1:(prop1,prop2,etc),prop3
Match 1:
    prop1
Match 2:
    subSubProp1:(prop1,prop2,etc)
    Group 1:
        subSubProp1
    Group 2:
        prop1,prop2,etc
Match 3:
    prop3

最后:

Input:
    prop1,prop2,etc
Match 1:
    prop1
Match 2:
    prop2
Match 3:
    etc

https://regex101.com/r/WAXrFd/2

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