Bash : extracting part of a string

Question

Say I have the string \"Memory Used: 19.54M\" How would I extract the 19.54 from it? The 19.54 will change frequently so i need to store it in a variable and compare it with

Answer 1

Other possible solutions:

With grep:

var="Memory Used: 19.54M"
var=`echo "$var" | grep -o "[0-9.]\+"`

With sed:

var="Memory Used: 19.54M"
var=`echo "$var" | sed 's/.*\ \([0-9\.]\+\).*/\1/g'`

With cut:

var="Memory Used: 19.54M"
var=`echo "$var" | cut -d ' ' -f 3 | cut -d 'M' -f 1`

With awk:

var="Memory Used: 19.54M"
var=`echo "$var" | awk -F'[M ]' '{print $4}'`

Answer 2

> echo "Memory Used: 19.54M" | perl -pe 's/\d+\.\d+//g'
Memory Used: M

Answer 3

You probably want to extract it rather than remove it. You can use the Parameter Expansion to extract the value:

var="Memory Used: 19.54M"
var=${var#*: }            # Remove everything up to a colon and space
var=${var%M}              # Remove the M at the end

Note that bash can only compare integers, it has no floating point arithmetics support.

Answer 4

You can use bash regex support with the =~ operator, as follows:

var="Memory Used: 19.54M"
if [[ $var =~ Memory\ Used:\ (.+)M ]]; then
    echo ${BASH_REMATCH[1]}
fi

This will print 19.54