Using grep -Po to extract regex

grep

I've been struggling with "grep -Po" off and on for a couple hours.

What I would like to do is to find out how many bytes the virtual disk size is as reported by qemu-img info.

Here's example output:

$ qemu-img info 022eb199-954d-4c78-8550-0ea1d31111ec                                           
image: 022eb199-954d-4c78-8550-0ea1d31111ec
file format: qcow2
virtual size: 40G (42949672960 bytes)
disk size: 20G
cluster_size: 65536

And here's what I've been trying variations of:

$ qemu-img info 022eb199-954d-4c78-8550-0ea1d31111ec | grep -Po '\(([0-9]+).*\)'
(42949672960 bytes)

But I would just like the number. I guess I'm struggling with how to tell "grep -Po" which part of the regex I'd like extracted. I thought the "(" brackets would do that.

Any help would be greatly appreciated. 🙂

Best Answer

You can use cut twice, for example:

echo "virtual size: 40G (42949672960 bytes)"| cut -f 4 -d" "| cut -c 2-

First cut cuts (42949672960,-f4 fourth field, -d" " is the separator. The second cut cuts 42949672960 -c2- (cuts from second character to the end.

Related Solutions

Ls | grep regex, ignore files with .tab,

How I would do this would depend on the context. If this wasn't for a script, and I just wanted to see this, I would be okay with using ls:

ls -a | egrep -v '(tab|\.o$|\.bin$|~$)'

If in a script, I would use a loop:

for file in *; do
#file $file in .* *; do   -- To Include dot (hidden files)                     
   if ! egrep -q '(tab|\.o$|\.bin$|~$)' <<<$file; then
      echo $file
   fi
done

You could add another if test with -f if you only want regular files and not directories. If you want to do this recursively, you could use GNU find with a -regex test and posix-egrep. But take into account the fact that this regex will match the full pathname relative to the directory you searched in.

Another nice option is extended globs if you have support for it, see Dennis's answer.

Grep date with regex

Brackets can be used to give you a range of numbers. The carrot can match the start of a line. So..

grep "^12\/1[6-8]\/09" will give you everything that starts with 12/16/09, 12/17/09, and 12/18/09

Best Answer

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Ls | grep regex, ignore files with .tab,

Grep date with regex

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