正则表达式获取捕获组的大小

是否可以编写一个正则表达式，以便我可以稍后在同一正则表达式中引用“第一个捕获组的长度”？我在这里想要实现的是捕获连续发生的情况1后面跟着连续出现的确切次数2's.

我想要类似的东西

r"(1*)(2{length(\1)})" # where `length(\1)` should give me the length of capture group 1

应该匹配

1122 # two 1's followed by two 2's
111222 # three 1's followed by three 2's
121122111222 # should match `12` and `1122` and `111222` separately

不应匹配

122 # there are two 2's following one 1
112 # there are two 1's but only one 2
11222 # same as above but with different occurrences
11122 # same as above but with different occurrences

Update我猜你可能会采用一些荒谬的 Java 前瞻递归模拟，但这是行不通的
或者你可以使用Python来做到这一点？

>>> import regex
>>> rx_1_2 = r"(?m)^(1(?>(?1))*2)$"
>>>
>>> input = '''
... 111222222
... 11222234
... 1111222
... 111222
... 1122
... 12
... '''
>>> res = regex.findall( rx_1_2, input )
>>> print( res )
['111222', '1122', '12']

该问题被标记为 Java 模拟递归的重复项
使用前瞻对于报道此内容的人来说是极其错误的判断 通过将其标记为重复来提出问题。简直就是判断力差...

可以用 python 来完成regex module.
需要用到递归。
这样做是因为它实际上只是嵌套的分隔符。

1
  1
    1
    2
  2
2

1(?>[^12]++|(?R))*2

https://regex101.com/r/4Nxtvl/1 https://regex101.com/r/4Nxtvl/1

                         # Recursion 
 1                       # 1
 (?>                     # Atomic group
      [^12]++                 # Possesive, not 1 or 2
   |                        # or,
      (?R)                    # Recurse the regex
 )*                      # End cluster, do 0 to many times
 2                       # 2

To 不允许内部内容 use 1(?>(?R))*2 https://regex101.com/r/mSUIp0/1 https://regex101.com/r/mSUIp0/1

要添加边界条件，请将递归包含到组中，
然后用边界结构包围它。

(?<!\d)(1(?>[^12]++|(?1))*2)(?!\d)

https://regex101.com/r/SSr1zV/1 https://regex101.com/r/SSr1zV/1

 (?<! \d )               # Not a digit behind
 (                       # (1 start), Recursion code group
    1                       # 1
    (?>                     # Atomic group
       [^12]++                 # Possesive, not 1 or 2
     |                        # or,
       (?1)                    # Recurse the regex group 1
    )*                      # End cluster, do 0 to many times
    2                       # 2
 )                       # (1 end)
 (?! \d )                # Not a digit ahead

To 不允许内部内容 use (?<!\d)(1(?>(?1))*2)(?!\d) https://regex101.com/r/VI6w0Y/1 https://regex101.com/r/VI6w0Y/1