.net RegEx：最小可能匹配或非贪婪匹配
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RegEx: Smallest possible match or nongreedy match
提问by Jonathan Allen
How do I tell RegEx (.NET version) to get the smallest valid match instead of the largest?
如何告诉 RegEx（.NET 版本）获得最小的有效匹配而不是最大的匹配？
回答by DMI
For a regular expression like .*or .+, append a question mark (.*?or .+?) to match as few characters as possible. To optionally match a section (?:blah)?but without matching unless absolutely necessary, use something like (?:blah){0,1}?. For a repeating match (either using {n,}or {n,m}syntax) append a question mark to try to match as few as possible (e.g. {3,}?or {5,7}?).
对于像.*or 之类的正则表达式.+，附加一个问号（.*?或.+?）以匹配尽可能少的字符。要可选地匹配某个部分(?:blah)?但除非绝对必要，否则不匹配，请使用类似(?:blah){0,1}?. 对于重复匹配（使用{n,}或{n,m}语法），附加一个问号以尝试匹配尽可能少的匹配（例如{3,}?或{5,7}?）。
The documentation on regular expression quantifiersmay also be helpful.
有关正则表达式量词的文档也可能有所帮助。
回答by dxh
The non-greedy operator, ?. Like so:
非贪婪运算符，?。像这样：
.*?
回答by Jonathan
The non greedy operator does not mean the shortest possible match:
非贪婪运算符并不意味着可能的最短匹配：
abcabk
abcbk
a.+?kwill match the entire string (in this example) instead of only the last three signs.
a.+?k将匹配整个字符串（在本例中），而不是仅匹配最后三个符号。
I'd like to actually find the smallest possible match instead.
我想实际上找到最小的匹配项。
That is that last possible match for 'a' to still allow all matches for k.
那是 ' a' 的最后一个可能匹配仍然允许k.
I guess the only way to do that is to make use of an expression like:
我想唯一的方法是使用这样的表达式：
a[^a]+?k
