Looking for an awk command to print strings only if substring is missing


 
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# 1  
Old 04-03-2014
Looking for an awk command to print strings only if substring is missing

I have a file that I need to find each interface that has move-group on the interface line and print this line if the lines under the interface does Not have "filter-shared 14".

Example file:
Code:
interface 1/1/1/0 move-group
     decription one one one zero
     no shut
     filter-shared 14
     accounting-policy 5
     shared-address 0
interface 1/1/1/1 move-group
     decription one one one one
     no shut
     accounting-policy 5
     shared-address 0
interface 1/1/1/2 move-group
     decription one one one two
     no shut
     filter-shared 14
     accounting-policy 5
     shared-address 0
interface 1/1/1/3 move-group
     decription one one one three
     no shut
     accounting-policy 5
     shared-address 0
interface 1/1/1/4 permanent-group
     decription one one one four
     no shut
     accounting-policy 5
     shared-address 0

I only wish to print out the interface line that does have "filter-shared 14" under it.
So the output would look like this:
Code:
interface 1/1/1/1 move-group
interface 1/1/1/3 move-group

# 2  
Old 04-03-2014
Does it have to be AWK? Try this if not:
Code:
perl -ln0e 'while (/(interface[^\n]+).*?(?=interface|$)/sg){$x=$&;$y=$1;print $y if $x!~/filter-shared 14/ && $y=~/move-group/}' file

This User Gave Thanks to bartus11 For This Post:
# 3  
Old 04-03-2014
Stash each interface line in a variable a, and 0 in another b, and as your seaarch forward, look for filter-shared 14, and if you find it, st the second var b to 1, and at the next interface line or EOF if b still a zero, print the stash a before stashing the new interface line. So, in code the last bit comes first!
# 4  
Old 04-03-2014
bartus11, That is a perfect perl line, but the server I need to run this on does not have perl.
# 5  
Old 04-03-2014
One way with awk:
Code:
awk '/^interface/{print x}1' file |
awk '!/filter-shared 14/ && $1~/move-group/{print $1}' FS='\n' RS=

This User Gave Thanks to Scrutinizer For This Post:
# 6  
Old 04-03-2014
Thank you Scrutinizer, that works. But it skips the first line of the file if there is a match. I think it's because x is empty until after the first line.
# 7  
Old 04-04-2014
Hi Numele, I cannot reproduce this skipping of the first line. Could you give an example and could you specify on what OS you are getting this result?

I get:
Code:
interface 1/1/1/1 move-group
interface 1/1/1/3 move-group

If I change "filter-shared 14" of the first record to something else, then the output becomes:
Code:
interface 1/1/1/0 move-group
interface 1/1/1/1 move-group
interface 1/1/1/3 move-group

Note that there are two parts. The first puts empty lines around the records, the "x" is an empty variable, so that an empty line gets printed before that start of each record.

This then enables the second awk to use RS=, a special value so that records can be separated by blank lines and thus become multi-line records. So for example the first record becomes:

Code:
interface 1/1/1/0 move-group
     decription one one one zero
     no shut
     filter-shared 14
     accounting-policy 5
     shared-address 0

Because the fields separator FS is set to \n, the first field ($1) is equal to the first line of such a multiline record..
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