用通配符匹配任何字符的正则表达式

I am new to regex and I am trying to come up with something that will match a text like below:

我是regex的新手，我正在尝试找到一些匹配如下文本的东西:

ABC: (z) jan 02 1999 \n

ABC: (z) 1999年1月02日\n

Notes:

注:

text will always begin with "ABC:"
文本总是以“ABC:”开头
there may be zero, one or more spaces between ':' and (z).
在':'和(z)之间可能有一个或多个空格。
Variations of (z) also possible - (zz), (zzzzzz).. etc but always a non-digit character enclosed in "()"
(z)的变化也可能——(zz)、(zzzzzz)…但总是包含在"()"中的非数字字符
there may be zero,one or more spaces between (z) and jan
在(z)和jan之间可能有一个或多个空格。
jan could be jan, january, etc
jan可以是jan, january等等
date couldbe in any format and may/may not contain other text as part of it so I would really like to know if there is a regex I can use to capture anything and everything that is found between '(z)' and '\n'
日期可以是任何格式，也可以/不可以包含其他文本作为它的一部分，所以我很想知道是否有一个regex可以用于捕获'(z)'和'\n'之间的所有内容

Any help is greatly appreciated! Thank you

非常感谢您的帮助!谢谢你！

3 个解决方案

#1

The following should work:

以下工作:

ABC: *\([a-zA-Z]+\) *(.+)

Explanation:

解释:

ABC:            # match literal characters 'ABC:'
 *              # zero or more spaces
\([a-zA-Z]+\)   # one or more letters inside of parentheses
 *              # zero or more spaces
(.+)            # capture one or more of any character (except newlines)

To get your desired grouping based on the comments below, you can use the following:

要根据下面的评论获得所需的分组，您可以使用以下方法:

(ABC:) *(\([a-zA-Z]+\).+)

#2

Without knowing the exact regex implementation you're making use of, I can only give general advice. (The syntax I will be perl as that's what I know, some languages will require tweaking)

如果不知道您正在使用的确切的regex实现，我只能给出一般的建议。(我将使用perl作为语法，因为我知道，有些语言需要进行调整)

Looking at ABC: (z) jan 02 1999 \n

看ABC: (z) 1999年1月02日

The first thing to match is ABC: So using our regex is /ABC:/
首先要匹配的是ABC:所以使用我们的regex是/ABC:/。
You say ABC is always at the start of the string so /^ABC/ will ensure that ABC is at the start of the string.

你说美国广播公司总是在字符串的开始/ ^校正/将确保在字符串的开始。
You can match spaces with the \s (note the case) directive. With all directives you can match one or more with + (or 0 or more with *)

您可以将空格与\s(注意到案例)指令相匹配。使用所有指令，您可以将一个或多个指令与+(或多个指令与*)匹配
You need to escape the usage of ( and ) as it's a reserved character. so 

您需要避免使用(and)，因为它是一个保留字符。所以
You can match any non space or newline character with .

您可以匹配任何非空格或换行字符。
You can match anything at all with .* but you need to be careful you're not too greedy and capture everything.

你可以和任何东西搭配。*但是你需要小心，你不是太贪心，什么都能捕捉到。

So in order to capture what you've asked. I would use /^ABC:\s*$.+?$\s*(.+)$/

为了抓住你的问题。我将使用/ ^ ABC:\ s * $+ ? $\ s *(. +)/美元

Which I read as:

我理解为:

Begins with ABC:

始于美国广播公司(ABC):

May have some spaces

可能有一些空间

has (

有(

has some characters

有一些字符

has )

)

may have some spaces

可能有一些空间

then capture everything until the end of the line (which is $).

然后捕获所有内容，直到行尾(即$)。

I highly recommend keeping a copy of the following laying about http://www.cheatography.com/davechild/cheat-sheets/regular-expressions/

我强烈建议保存以下关于http://www.cheatography.com/davechild/cheat-sheets/regular-expressions/的文章

#3

This should fulfill your requirements.

这应该满足您的要求。

ABC:\s*($\D+$\s*.*?)\\n

美国广播公司(ABC):\ s *(\ \ D + \ \ s *。* ?)\ \ n

Here it is with some tests http://www.regexplanet.com/cookbook/ahJzfnJlZ2V4cGxhbmV0LWhyZHNyDgsSBlJlY2lwZRiEjiUM/index.html

这里有一些测试http://www.reg解释器.com/cookbook/ahjzfnjlz2v4cgxhbmv0lwhyzhnydgbljly2lwzriejium/index.html

Futher reading on regular expressions: http://www.regular-expressions.info/characters.html

进一步阅读正则表达式:http://www.regular-expressions.info/characters.html

#1