How do I ignore forward slash and space at the start of the line in regular expressions?
In the Example below, I need to ignore the pipe and space because I am using grep and awk
The actual command gives me
cmd
size=5.0G features='0' hwhandler='0' wp=rw
|-+- policy='round-robin 0' prio=1 status=active
| `- 3:0:0:3 sdh 8:112 active ready running #Line 3
`-+- policy='round-robin 0' prio=1 status=enabled
`- 4:0:0:3 sdl 8:176 active ready running #Line 5
By doing this:
cmd | grep -E '[0-9]+:[0-9]+:[0-9]+:[0-9]+' | awk '{print $3}'
I was able to get the sdh, sdl. But the problem is, I need to ignore the '|' upfront, to make the Line 3 and Line 5 same. Please advise.
Edit 1 I need to get two information
1) the Number
3:0:0:3
4:0:0:3
2) Disk name
sdh
sdl
Rather than try to make each of your two lines have identical number of fields, just use the -o option of grep to only part of the line that matches your regular expression. Then you won't need the awk command at all.
cmd | grep -o -E '[0-9]+:[0-9]+:[0-9]+:[0-9]+'
Since you actually need more than just what was in your original question:
cmd | grep -E '[0-9]+:[0-9]+:[0-9]+:[0-9]+' | sed 's/^| //' | awk '{print $2, $3}'