search and retrieve previous line in file

paulsew · #1 (**permalink**) 02-23-2007

I've got a file with the following layout:

#STMP FSgroup
filename /filesysname1
filestatus 2

#STMP FSstatus
filename /filesysname1
ratio 30

#STMP FSgroup
filename /filesysname1
filestatus 2

#STMP FSstatus
filename /filesysname6
ratio 60

I'm trying to write a SHELL (ksh) script that'll search the file say e.g filename = /filesysname1 grouped under FSstatus and get everthing beneath FSstatus group.

Thus, I should be able search for anygroup, i.e. lines that start with an # and everything below it until the first blank line. I also need to know the group name, i.e. either FSstatus or FSgroup

Look really difficult for me, any help out there

tks
William

matrixmadhan · #2 (**permalink**) 02-23-2007

Will this be ok?

Code:

var=filesysname1
awk -F" " 'BEGIN{group=""; x=0} { if( x == 1 ) { print } if( $0 ~ /^#/) { group = $0; next } if( $0 ~ /'$var'/ ) { x=1; print group; print $0} else { x = 0 } }' file

anbu23 · #3 (**permalink**) 02-23-2007

Quote:

I'm trying to write a SHELL (ksh) script that'll search the file say e.g filename = /filesysname1 grouped under FSstatus and get everthing beneath FSstatus group.

If all the groups are separated by blank line you can use this

Code:

awk -v RS="" -v FS="\n" -v ORS="\n\n" ' $1 ~ "FSstatus" && $0 ~ "/filesysname1" ' file

Quote:

Thus, I should be able search for anygroup, i.e. lines that start with an # and everything below it until the first blank line. I also need to know the group name, i.e. either FSstatus or FSgroup

If all the groups are separated by blank line you can use this to display group with group name "FSstatus"

Code:

awk -v RS="" -v FS="\n" -v ORS="\n\n" ' $1 ~ "FSstatus" ' file

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