在bash中找到一行中的特定字符

Question

Is there a way to extract a particular part from a line of text with regular expression in bash.

For example

8495 pts/1 S 0:10 /opt/sbin/proc -q 10 -p /opt/etc/plugin/ -c /opt/etc/ -f 200000 -s 10000 -b 3276

How can I extract q parameter 10 from above?. I need the number after q parameter

Answer 1

px aux | sed -n 's/.*-q\s\([0-9]\+\).*/\1/p'

Answer 2

The naive solution regex would be .*-q ([^ ]*) .* . So, something like sed 's/.*-q ([^ ]*) .*/\\1/' should do the trick.

Answer 3

If the line is stored in variable $line :

val=$(
    this_token=false 
    for word in $line; do 
        $this_token && { echo $word; break; } 
        [[ $word = "-q" ]] && this_token=true 
    done
) 
echo $val