正则表达式 – Powershell:从字符串中提取文本

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如何从字符串中提取“程序名称”。字符串将如下所示:

% O0033(SUB RAD MSD 50R III) G91G1X-6.4Z-2.F500 G3I6.4Z-8. G3I6.4
G3R3.2X6.4F500 G91G0Z5. G91G1X-10.4 G3I10.4 G3R5.2X10.4 G90G0Z2.
M99 %

程序名称为(SUB RAD MSD 50R III)。将结果存储在另一个字符串中是正确的。我正在学习powerhell,所以任何解释你的答案将不胜感激。

以下正则表达式在括号之间提取任何内容
PS> $prog = [regex]::match($s,'\(([^\)]+)\)').Groups[1].Value
PS> $prog
SUB RAD MSD 50R III


Exlanation (created with RegexBuddy)

Match the character '(' literally «\(»
Match the regular expression below and capture its match into backreference number 1 «([^\)]+)»
   Match any character that is NOT a ) character «[^\)]+»
      Between one and unlimited times,as many times as possible,giving back as needed (greedy) «+»
Match the character ')' literally «\)»

检查这些链接

@L_502_0@

http://powershell.com/cs/blogs/tobias/archive/2011/10/27/regular-expressions-are-your-friend-part-1.aspx

http://powershell.com/cs/blogs/tobias/archive/2011/12/02/regular-expressions-are-your-friend-part-2.aspx

http://powershell.com/cs/blogs/tobias/archive/2011/12/02/regular-expressions-are-your-friend-part-3.aspx

原文链接:https://www.f2er.com/regex/357360.html

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