AWK get specificic pattern - awk

Question

I have lines like this:
Volume.Free_IBM_LUN59_28D: 2072083693568
I would like to get only IBM_LUN59_28D from this line using awk.
Thanks

Answer 1

You can use sub to do substitutions on each input line, as per the following transcript:
pax> echo 'Volume.Free_IBM_LUN59_28D: 2072083693568' | awk '
...> {
...> sub (".*Free_", "");
...> sub (":.*", "");
...> print
...> }'
IBM_LUN59_28D
That command crosses multiple lines for readability but, if you're operating on a file and not too concerned about readability, you can just use the compressed version:
awk '{sub(".*Free_","");sub(":.*","");print}' inputFile
If you're amenable to non-awk solutions, you could also use sed:
sed -e 's/.*Free_//' -e 's/:.*//' inputFile
Note that both those solutions rely on your (somewhat sparse) test data. If your definition of "like" includes preceding textual segments other than Free_ or subsequent characters other than :, some more work may be needed.
For example, if you wanted the string between the first _ and the first :, you could use:
awk '{sub("[^_]*_","");sub(":.*","");print}'

Answer 2

With sed:
sed 's/[^_]*_\(.*\):.*/\1/'
Search for sequence of non _ characters followed by _ (this will match Volume.Free_), then another sequence of characters (this will match IBM_LUN59_28D, we group this for future use), followed by : and any char sequence. Substitute with the saved pattern (\1). That's it.
Sample:
$ echo "Volume.Free_IBM_LUN59_28D: 2072083693568" | sed 's/[^_]*_\(.*\):.*/\1/'
IBM_LUN59_28D

Answer 3

Here is one awk
awk -F"Free_" 'NF>1{split($2,a,":");print a[1]}'
Eks:
echo "Volume.Free_IBM_LUN59_28D: 2072083693568" | awk -F"Free_" 'NF>1{split($2,a,":");print a[1]}'
IBM_LUN59_28D
It divides the line by Free_.
If line then have more than one field NF>1 then:
Split second field bye : and print first part a[1]

Answer 4

With awk:
echo "$val" | awk -F: '{print $1}' | awk -F. '{print $2}' | awk '{print substr($0,6)}'
where the given string is in $val.