From: Hustert Klaus-CKH035 (Klaus.Hustert@motorola.com)
Date: Tue Aug 19 2003 - 03:40:44 EDT
Hi folks,
Thanks a lot to all for such amount of helpfully answers!
#----------------------------------------------------------------------
bpmedialist | awk ' !NF == 0 && !/Server Host =/ && !/images/ && !/---------/ { print $0 }' | \
awk '
{ if ( (NR % 2 ) == 0 )
printf"%s%s\n", prev_line,$0
else
prev_line = $0" " }
Thx 2 my colleague Jim McDonald ( I did not know about him before ;-). Hi Jim.
#----------------------------------------------------------------------
Assuming you have the input in file /tmp/in
The following command will give you every line _before_ the line that ends with "FROZEN"
sed -n '/FROZEN$/{;g;p;}
h;' /tmp/in
(I like sed ;-)
Thx 2 Harm
#----------------------------------------------------------------------
Long_command | sed -n '/FROZEN/{
x
p
x
p
}
h'
Thx 2 Alan
#----------------------------------------------------------------------
I suggest you use the 'available_media' script in the /usr/openv/netbackup/bin/goodies directory.
Thx 2 Tim
#----------------------------------------------------------------------
Try this code:
BEGIN {
# train awk to use multi-line records
RS=""
FS="\n"
}
/FROZEN/ {
# since whole "line" is now a "field", we will have
# to break-up field one on white space to get the id
split($1,junk," ")
print junk[1]
}
Thx 2 David
#----------------------------------------------------------------------
Here the answers and on bottom my question:
awk '{if(substr($1,1,2) == "BE") { id=$1; getline ; if($NF == "FROZEN") print id}}'
This will print only the ID if you have FROZEN as the status. If you want the whole line rather than just the ID, then replace the "id = $1" with "id = $0". If you want both lines if the status is frozen, do as mentioned above, but also instead of "print id" od "print $0".
Thx 2 Vince
#----------------------------------------------------------------------
In awk, you can set the end-of-line to be a blank line, and then the fields will be numbered from 1 - N between blank lines:
##
# joanna:~stuart/awk/patsup.awk
#
#
# Variables:
# ARGC Argument Count
# ARGV Array of command line arguments
# FILENAME Input filename
# FNR Current record number
# FS Field seperator
# NF Numver of fields
# NR Current record number
# OFS Output field Spererator
# ORS oupt record seperaotr
# RS (input) record seperator
# $1 First field
# $0 entire line
BEGIN {
RS = "" # set record seperator to blank line
}
/^BE0/ {
if ($13 == "FROZEN") print $0
}
END {
print "Done!"
}
Something like that...
Thx 2 Stuart
#----------------------------------------------------------------------
/usr/openv/netbackup/bin/admincmd/bpmedialist | awk ' BEGIN { ###Print the heading lines if ($1 == "id")
print $0
if ($1 == "vimages")
print $0
if (substr($1,1,5) == "-----")
print $0
###Save the BE line for later pringting.
if (substr($1,1,2) == "BE")
line_one = $0
###Print the lines
if ($4 == "FROZEN") {
print line_one
print $0
}
}'
Thx 2 Ronelle
#----------------------------------------------------------------------
awk '! /FROZEN/ {l=$*} /FROZEN/ {print $l; print $*}'
this will print all lines containing FROZEN and the immediately preceeding one ("l" acts as a buffer)
you could also try the GNU version of grep which allows you to print context lines
grep -B1 FROZEN
which says : print all lines containing FROZEN and one line before
Thx 2 Thierry
#----------------------------------------------------------------------
bash-2.03# /usr/openv/netbackup/bin/admincmd/bpmedialist | awk '$1 ~ /BE[0-9][0-9]*/ {
getline
if($NF == "FROZEN") print}'
}'
Thx 2 Sundar
#----------------------------------------------------------------------
awk '{a=$0;getline; b=$0;if ( b ~ /FROZEN/) {print a; print b;}; getline}' a.txt
Thx 2 balaji
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