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Full Discussion: grep and cut problem
Top Forums UNIX for Dummies Questions & Answers grep and cut problem Post 302217930 by zaxxon on Thursday 24th of July 2008 12:37:11 AM
Old 07-24-2008
I guess I live in another timezone and when work is over, I go home and usually don't look into here anymore. That's why I didn't answer that fast Smilie Can you post the pm question here please? I will describe the syntax soon as edit in this post here.

EDIT:

Code:
sed 's/.*<TAG1>\([^>]*\)<\/TAG1>.*/\1/g'

s = substitution
/ = start of the pattern
. = any character
* = zero or as many of the former character
\( = escaped starting bracket of the group I want to extract
[ = starting squared bracket of a group of characters
^ = inside the squared bracket means "not/none" of the following, in this case as long as no > shows up
] = ending the group
* = zero or as many of the former character, in this case as many as it is no >
\) = escaped bracket to end the group definition
\/ = just escaping the slash of that end tag
.* = zero or as many of any character (you know that by know already Smilie )
/ = here ends the pattern I want to find and starts that, through what I want to substitute
\1 = print out the 1st group I defined inside the \( \) escaped curled brackets
/ = end of the substitution input
g = globally, do it on the whole line of input


Best search the web for sites that explain regular expressions (reg exp) or get that nice awk&sed book from O'Reilly which is worth it, the small reference book and/or the "bigger" one.

Last edited by zaxxon; 07-24-2008 at 01:48 AM..
 

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