[英]C# Regex - only match if substring exists?
Ok, so I think I've got a handle on negation - now what about only selecting a match that has a specified substring within it?好的,所以我想我已经掌握了否定的方法 - 现在只选择一个包含指定子字符串的匹配怎么办?
Given:鉴于:
This is a random bit of information from 0 to 1.
This is a non-random bit of information I do NOT want to match
This is the end of this bit
This is a random bit of information from 0 to 1.
This is a random bit of information I do want to match
This is the end of this bit
And attempting the following regex:并尝试以下正则表达式:
/(?s)This is a random bit(?:(?=This is a random).)*?This is the end/g
Why isn't this working?为什么这不起作用? What am I missing?
我错过了什么?
I'm using regexstorm.com for testing...我正在使用 regexstorm.com 进行测试...
You ruined a tempered greedy token by turning the negative lookahead into a positive one.你把消极的前瞻变成了积极的前瞻,这毁了一个温和的贪婪令牌。 It won't work that way because the positive lookahead requires the text to equal
This is a random
at each position after This is a random bit
.它不会那样工作,因为正向前瞻要求文本等于
This is a random
at each position after This is a random bit
。
You need:你需要:
This is a random bit
)This is a random bit
)This is a random
)This is a random
)This is the end
)This is the end
) So, use所以,使用
(?s)This is a random bit(?:(?!This is a random bit|This is the end|This is a random).)*This is a random(?:(?!This is a random bit|This is the end).)*This is the end
See the regex demo查看正则表达式演示
(?s)
- DOTALL mode on ( .
matches a newline) (?s)
- DOTALL 模式 ( .
匹配换行符)This is a random bit
- Leading delimiter This is a random bit
- 前导分隔符(?: # Start of the tempered greedy token (?!This is a random bit # Leading delimiter | This is the end # Trailing delimiter | This is a random) # Sepcific string inside . # Any character )* # End of tempered greedy token
This is a random
- specified substring This is a random
指定的子串(?:(?!This is a random bit|This is the end).)*
- Another tempered greedy token matching any text not leading/closing delimiters up to the first... (?:(?!This is a random bit|This is the end).)*
- 另一个缓和的贪婪标记匹配任何文本,直到第一个前/结束分隔符...This is the end
- trailing delimiter This is the end
尾随定界符I hope you understand this (?:(?=This is a random).)
can only match once, never twice if it were quantified.我希望你明白这一点
(?:(?=This is a random).)
只能匹配一次,如果被量化则永远不会匹配两次。 For example Th
can satisfy the lookahead.例如
Th
可以满足前瞻。 When the T
is consumed, the next character is h
which will never satisfy the lookahhead Th
.当
T
被消耗时,下一个字符是h
,它永远不会满足 Lookahhead Th
。 The next expression is evaluated, never to return to the lookahead again.评估下一个表达式,永远不会再次返回前瞻。 Use a negative lookahead instead.
改用负前瞻。
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