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Sed: printing lines AFTER pattern matching EXCLUDING the line containing the pattern

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Top Forums Shell Programming and Scripting Sed: printing lines AFTER pattern matching EXCLUDING the line containing the pattern
# 8  
Old 06-04-2014
All you need is egrep -A piped into grep -v. Set the -A for the number of rows that you want to see after the row you are looking for and use grep -v to filter the row that you were looking for. you don't need sed or a for loop.

Code:
$ egrep -A1 "alias top" .bashrc
alias top='xtitle Processes on $HOST && top'
alias make='xtitle Making $(basename $PWD) ; make'

$ egrep -A1 "alias top" .bashrc | grep -v "alias top"
alias make='xtitle Making $(basename $PWD) ; make'

# 9  
Old 06-05-2014
My server doesn't support -A on grep. Besides, the statement that I've written is already a complex one and I want to use sed as part of multiple piped commands. Using 3-4 lines of code there would not be feasible.

Quote:
Originally Posted by rbatte1
Does your server support the -A flag on grep? That might be a neater solution:-
Code:
grep -A10 pattern  file

Maybe I've missed the point. If you want all lines after a certain point, perhaps you could use grep to get the record number and then tail to get you the output you need:-
Code:
#!/bin/ksh
 
grep -n pattern  file|tr ":" " "|read line xxx       # Get the interesting line number
wc -l file | read total xxx                          # Get total lines
((readrecs=$total-$line))                            # Work out how many lines from end to read
tail -$readrecs   file


Do either of these help?


Robin
---------- Post updated at 02:04 AM ---------- Previous update was at 01:51 AM ----------

Made it work somehow.. although still looking for a better solution. Here's what I am using for the time being:

Code:
sed -n '/Conn.*User/,${//!p;}'|sed '1d'

# 10  
Old 06-05-2014
Quote:
Originally Posted by essem
[..]

Made it work somehow.. although still looking for a better solution. Here's what I am using for the time being:

Code:
sed -n '/Conn.*User/,${//!p;}'|sed '1d'

So have the requirements changed? This will remove the line that contains the pattern plus the line after that line, so effectively the first two lines.....
# 11  
Old 06-05-2014
The first sed command searches for the pattern and prints the lines after the pattern match (including the line that contains the pattern).

The second sed command (after pipe) deletes the first line from the output of the first sed command.

The requirement is still the same. This is just a workaround until I find a real fix.

Quote:
Originally Posted by Scrutinizer
So have the requirements changed? This will remove the line that contains the pattern plus the line after that line, so effectively the first two lines.....
# 12  
Old 06-05-2014
No, the first sed command (sed -n '/Conn.*User/,${//!p;}') prints the lines after the pattern match excluding the line(s) that contain the pattern.

What is your OS and version?
Last edited by Scrutinizer; 06-05-2014 at 06:08 AM..
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