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Top Forums UNIX for Dummies Questions & Answers Deleting words between every appearance of two words Post 302918524 by pilnet101 on Wednesday 24th of September 2014 01:57:02 AM
Old 09-24-2014
As per your exact requirements, a space would remain between > and FMG or GFT.

The below sed command will produce the correct output:

Code:
sed 's/\(^.*\)gi.*Fesd\(.*$\)/\1\2/' file

To remove the space and get the output which you have shown you can amend the above command to the below:

Code:
sed 's/\(^.*\)gi.*Fesd[[:space:]]\(.*$\)/\1\2/' file

This User Gave Thanks to pilnet101 For This Post:
 

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look(1) 							   User Commands							   look(1)

NAME
look - find words in the system dictionary or lines in a sorted list SYNOPSIS
/usr/bin/look [-d] [-f] [-tc] string [filename] DESCRIPTION
The look command consults a sorted filename and prints all lines that begin with string. If no filename is specified, look uses /usr/share/lib/dict/words with collating sequence -df. look limits the length of a word to search for to 256 characters. OPTIONS
-d Dictionary order. Only letters, digits, TAB and SPACE characters are used in comparisons. -f Fold case. Upper case letters are not distinguished from lower case in comparisons. -tc Set termination character. All characters to the right of c in string are ignored. FILES
/usr/share/lib/dict/words spelling list ATTRIBUTES
See attributes(5) for descriptions of the following attributes: +-----------------------------+-----------------------------+ | ATTRIBUTE TYPE | ATTRIBUTE VALUE | +-----------------------------+-----------------------------+ |Availability |SUNWesu | +-----------------------------+-----------------------------+ SEE ALSO
grep(1), sort(1), attributes(5) SunOS 5.10 29 Mar 1994 look(1)
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