How can I replace a / with a _ in a shell pipeline?

Question

Running

ip -o -f inet addr show | grep $INTERNAL |awk '/scope global/ {print $4}'

Want to replace the / in my output to _ so rather than reading

10.168.122.59/16

it reads as

10.168.122.59_16

.

|sed s///_/

didnt help

Any suggestions?

Can do it all in awk `ip -o -f inet addr show | awk -vI="$INTERNAL" '$0~I&&/scope global/{sub(/\//,"_",$4);print $4}'` — 123, May 14 '17 at 20:02
@CharlesDuffy You can if you want, pretty sure this is definitely a duplicate though. — 123, May 14 '17 at 21:56
Hmm. There's definitely a case for http://stackoverflow.com/questions/5864146/use-slashes-in-sed-replace, though that one assumes `sed`. — Charles Duffy, May 14 '17 at 22:13

score 2 · Accepted Answer · edited May 23 '17 at 12:26

All the postprocessing requested can be done internal to awk. Expanding a one-liner provided in a comment by @123 for better readability, this can look like the following:

ip -o -f inet addr show | \
  awk -v i="$INTERNAL" '
    $0 ~ i && /scope global/ { 
      sub(/\//, "_", $4);
      print $4;
    }'

Breaking down how this works:

awk -v i="$INTERNAL" defines an awk variable based on a shell variable. (As an aside, all-caps shell variable names are bad form; per POSIX convention, lowercase names are reserved for application use, whereas all-caps names can have meaning to the OS and surrounding tools).
$0 ~ i filters for the entire line ($0) matching the awk variable i.
/scope global/ by default is applied as a regex against $0 as well (it's equivalent to $0 ~ /scope global/).
sub(/\//, "_", $4) substitutes /s with _s in the fourth field.
print $4 prints that field.

score 1 · Answer 2 · answered May 14 '17 at 21:36

1

You need to scape the / or use a different separator as below:

echo 10.168.122.59/16 | sed s:/:_:

answered May 14 '17 at 21:36

Carles Fenoy

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If you're going to use `sed`instead of `tr`, at least use `sed y:/:_:` – William Pursell May 18 '17 at 22:47

score 0 · Answer 3 · answered May 18 '17 at 22:39

0

echo 10.168.122.59/16| awk '{sub(/\//,"_")}1'
10.168.122.59_16

answered May 18 '17 at 22:39

Claes Wikner

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How can I replace a / with a _ in a shell pipeline?

3 Answers3