在Linux上的C中使用正则表达式 [英] using regular expressions in C on linux

问题描述

团队，
您能告诉我这种特殊模式的含义吗?
我只是一个初学者，在一个网站中发现了这个例子
通过regex.h提供reg表达的示例

Hi Team,
Can you please tell what is the meaning of this particular pattern.
I am just being a beginner and found this as an example in one of the site
providing example for reg expression through regex.h

char *PatternSearch = "[?&]XYZ=\\([[:alnum:]]*\\)";

如果您可以通过示例解释此模式的目的，请感谢您的答复


问候

Appreciate your response if you could explain what does this pattern look for with an example


Regards,

推荐答案

[?&]是?或&
之一 XYZ=就是那个精确的文本(区分大小写)
\\(是左括号字符(. \\被编译器转义，因此它将\(传递给正则表达式
[:alnum:]是A-Za-z0-9的简写(无方括号)，扩展为ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789.请注意，[]是此扩展的一部分.
[[:alnum:]]搜索[A-Za-z0-9]
[[:alnum:]]*搜索任意数量的[[:alnum:]](请参阅上一个)
\\)是右括号).再次\\被编译器转义，将\(作为正则表达式传递给

因此，它正在搜索

[?&] is one of ? or &
XYZ= is just that exact text (case sensitive)
\\( is the opening bracket character (. The \\ is escaped by the compiler, so it is passing \( into the regex
[:alnum:] is shorthand for A-Za-z0-9 (no square brackets) which expands to ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789. Note that the [] is part of this expansion.
[[:alnum:]] searches for [A-Za-z0-9]
[[:alnum:]]* searches for any number of [[:alnum:]] (see previous)
\\) is the closing bracket character ). Again \\ is escaped by the compiler, passing \( in as the regex

So, it is searching for

?XYZ=(/*any characters in here*/)

或

or

&XYZ=(/*any characters in here*/)

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