case

[ozi403]

Hi,

My output like this :

name: odin;
browse policy: Year;
retention policy: Year;
group: ;

name: odin;
browse policy: Month;
retention policy: Year;
group: ;

So I want :
If group: is ";", don't show name, browse policy and retention policy.
Thanks for helps.

 

[vgersh99]

nawk -f ozi.awk myOutput

ozi.awk:

BEGIN {
  RS=FS=""
  ORS="\n\n"
}
$NF !~ /[ ][ ]*group:[ ][ ]*;$/

vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[feherke]

Hi

Supposing that the sections are delimited with a blank line, this [tt]bash[/tt] script will read the file and store the lines until encounters a section limit. Then if anywhere in the last section was a group line with a value, then print the entire section.

sect=""; ok=0
while read str; do
  if test -z "$str"; then
    test $ok -eq 1 && echo "$sect"
    sect=""; ok=0
  else
    sect="$sect$str"$'\n'
    expr "$str" : "group: .\+;" > /dev/null && ok=1
  fi
done < datafile.txt

Feherke.

http://rootshell.be/~feherke/

 

[feherke]

Hi

With [tt]sed[/tt] :

sed -f 'H;/^$/{g;s/\n//;/group: .\+;/p;s/.*//;h}' datafile.txt

Feherke.

http://rootshell.be/~feherke/