非贪婪定期EX pression比赛在AWK分隔符多字符 [英] Non-greedy regular expression match for multicharacter delimiters in awk

问题描述

考虑字符串AB 1 BA 2 AB 3 BA。我怎么能AB和在非贪婪时尚BA（在AWK）之间的内容相匹配？

Consider the string "AB 1 BA 2 AB 3 BA". How can I match the content between "AB" and "BA" in a non-greedy fashion (in awk)?

我曾尝试以下内容：

awk '
BEGIN {
    str="AB 1 BA 2 AB 3 BA"
    regex="AB([^B][^A]|B[^A]|[^B]A)*BA"
    if (match(str,regex))
        print substr(str,RSTART,RLENGTH)
}'

无输出。我相信不匹配的原因是，有AB和BA之间的奇数个字符。如果我更换 STR 与AB 11 BA 22 AB 33 BA正则表达式似乎工作..

with no output. I believe the reason for no match is that there is an odd number of characters between "AB" and "BA". If I replace str with "AB 11 BA 22 AB 33 BA" the regex seems to work..

推荐答案

合并你的两个否定的字符类和删除 [^ A] 从第二交替：

Merge your two negated character classes and remove the [^A] from the second alternation:

regex = "AB([^AB]|B|[^B]A)*BA"

这正则表达式的字符串失败亚的斯亚贝巴，虽然 - 不知道这是一个问题。

This regex fails on the string ABABA, though - not sure if that is a problem.

说明：

AB       # Match AB
(        # Group 1 (could also be non-capturing)
 [^AB]   # Match any character except A or B
|        # or
 B       # Match B
|        # or
 [^B]A   # Match any character except B, then A
)*       # Repeat as needed
BA       # Match BA

因为只有这样，才能在交替匹配 A 是之前除了 B 匹配的字符，我们可以安全地使用简单的 B 作为备选方案之一。

Since the only way to match an A in the alternation is by matching a character except B before it, we can safely use the simple B as one of the alternatives.

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