*, ?, + natures all meancompet this nature. 什么性质意味着没有达到这一目的? 实例将有所帮助。
您可以使用否定的特性类别来排除某些特性:例如[^abcde],除一个、b、c、d、e等特性外,还将与任何特性相对应。
除字面上注明所有特性外,您可使用shorthands内称职:[w](downcase)与任何“序号”(字母、编号和强调)相匹配,[W](uppercase)将与字体相匹配;同样,[d]>将与0-9位数相匹配,而[D]<><>>>>>>>>除0-9位数外,对等。
如果你使用PHP,你可以查阅.regex nature阶层文件。
说“高端对口”有两种方式:特性范围,零宽的负面 look头/l头。
前者:无与<代码>a,b,c或0: [^a-c0]
后者:除<代码>foo和bar外,与任何三字母相匹配:
(?!foo/>,{3}
或
{3}(?<!foo/bar)
Also, a c或rection f或 you: *, ? and + do not actually match anything. They are repetition operat或s, and always follow a matching operat或. Thus, a+ means match one 或 m或e of a, [a-c0]+ means match one 或 m或e of a, b, c 或 0, while [^a-c0]+ would match one 或 m或e of anything that wasn t a, b, c 或 0.
[^] (in [ ] >) 是常规表述中的否定,而^<>>>>是“援引”的。
<> 代码>[^a-z]对从“a”到“z”的任何单一特性进行对比。
^[a-z]系指从“a”到“z”的起步。
https://en.wikipedia.org/wiki/Regular_expression”rel=“noreferer”>Reference<
^ used at the beginning of a character range, or negative lookahead/lookbehind assertions.
>>> re.match( [^f] , foo )
>>> re.match( [^f] , bar )
<_sre.SRE_Match object at 0x7f8b102ad6b0>
>>> re.match( (?!foo)... , foo )
>>> re.match( (?!foo)... , bar )
<_sre.SRE_Match object at 0x7f8b0fe70780>
Recently I discovered two amazing regular expression features: ?: and ?!. I was curious of other neat regex features. So maybe you would like to share some tricky regular expressions.
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