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 the value on the next iteration.

I imagine I need some combination of grep and regex, but I never really understood regex..

possible duplicate of [How to trim whitespace from bash variable?](http://stackoverflow.com/questions/369758/how-to-trim-whitespace-from-bash-variable) — user2284570, Jul 26 '14 at 14:10

score 83 · Accepted Answer · answered Oct 01 '12 at 10:25

83

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.

answered Oct 01 '12 at 10:25

choroba

216,930
22
195
267

Extract does better explain what I want to do. How then would I change the third line to remove the "." and all that follows it? – confusified Oct 01 '12 at 10:27
10

`${var%.*}` removes a dot and anything following it at the end of the string. – choroba Oct 01 '12 at 10:38
may i ask what does # and * here means? – once Apr 01 '16 at 10:20
3

@once: As written in the comments, `#` means "remove from the left", and `*` is just a wildcard that matches everything. – choroba Apr 01 '16 at 15:45

score 47 · Answer 2 · edited Oct 02 '12 at 11:38

47

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}'`

edited Oct 02 '12 at 11:38

Littm

4,903
4
29
37

answered Oct 01 '12 at 15:09

Onilton Maciel

3,419
1
21
28

Alastair · Answer 3 · 2016-03-24T21:39:11.930

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

edited Mar 24 '16 at 21:39

answered Jan 11 '16 at 18:46

Alastair

4,325
1
25
23

best BASH parser! – ir0h Jun 17 '21 at 22:50

score 1 · Answer 4 · edited Oct 01 '12 at 13:34

1

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

edited Oct 01 '12 at 13:34

glenn jackman

223,850
36
205
328

answered Oct 01 '12 at 10:43

Vijay

62,703
87
215
314

Bash : extracting part of a string

4 Answers4

Linked

Related