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# 1  
Old 01-19-2009
Search this display these...
Hi i would like to search for a pattern and then prints the 2 lines before the line where the pattern is found.
I would also like to apply this in compressed files.
Please help me in this.
# 2  
Old 01-19-2009
If you are using perl, you can use a while loop to read STDIN, and say set a line variable at 2. Then use an array for preceding lines, push to that until the count is greater then the line variable, use an if statement.

Then in the main section, once the regex is found (use if) you can dump out the array and line with the match.
# 3  
Old 01-19-2009
im not using perl

any ideas if i can do this using sed / awk or grep?
# 4  
Old 01-19-2009
HP-UX i think is the one im using
# 5  
Old 01-19-2009
Hammer & Screwdriver One possible approach
This shows how to get the line before a matching entry; you can change some of the fields to match up to your requirements exactly.

First I show a four-line sample file
Then I determine the line number that has the text in it
Lastly, I display the result; tail -1 means only one line. If I wanted to display two lines, use tail -2


Code:
> cat file143
program 1 - program to do this
program 2 - program to do that - and the other
program 3 - program to do nothing
program 4 - program that wishes - it would do something
> linn=`cat -n file143 | grep "nothing" | awk '{print $1}'`
> head -`echo ${linn} -1|bc` file143 | tail -1
program 2 - program to do that - and the other

# 6  
Old 01-19-2009
i cant get this to work
is this a one line command only?
nothing happens when i run it
# 7  
Old 01-19-2009
it seems that nothiong really happens here
linn=`cat -n file143 | grep "nothing" | awk '{print $1}'`

but this displays the results
linn=`cat -n file143 | grep "nothing" | awk '{print $1}'`


the problem is
it only displays the first occurence
how do i display all occurence?
 
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