multiple strings replacements on variable

Question

does bash support multiple strings replacement on a variable at once?

on a variable (example V variable) I want to replace XXXXXXX with a string1, YYYYY with a string2 and ZZZZZZZ with a string3

Is it possible to run 1 such relpacement command instead of the 3 run below?

$ V="AAAAAAA/XXXXXXX/BBBBBB/YYYYY/CCCCCC/ZZZZZZZ"
$ V=${V//XXXXXXX/string1}
$ V=${V//YYYYY/string2}
$ V=${V//ZZZZZZZ/string3}
$ echo $V
AAAAAAA/string1/BBBBBB/string2/CCCCCC/string3
$

Tripp Kinetics · Answer 1 · 2018-04-03T19:49:57.883

2

It's easier if you use sed.

V=$(echo $V | sed -e 's/XXXXXXX/string1/' -e 's/YYYYY/string2/' -e 's/ZZZZZZZ/string3/')

edited Apr 03 '18 at 19:49

answered Apr 03 '18 at 19:46

Tripp Kinetics

5,178
2
23
37

There are missing slashes (after string2 and after string3) – ErikMD Apr 03 '18 at 19:47
I got `sed: -e expression #2, char 15: unterminated 's' command` – Arkadiusz Drabczyk Apr 03 '18 at 19:47
Try it now. Forgot the terminating slashes – Tripp Kinetics Apr 03 '18 at 19:48
Now it's ok..... – Arkadiusz Drabczyk Apr 03 '18 at 19:48

score 1 · Accepted Answer · answered Apr 03 '18 at 19:48

Yes you can do it like below:-

V=$(echo "$V" | sed 's/XXXXXXX/string1/g;s/YYYYY/string2/g;s/ZZZZZZZ/string3/g')

How will it work? sed command will search for sting XXXXXX and replace it with string2, replace string YYYYYY with string2 and finally replace ZZZZZZZ with string3 and store the new string into variable V.

So now check the result like below:-

echo $V

ErikMD · Answer 3 · 2018-04-03T19:59:12.127

1

@Abhijit's answer is very OK (the sed option /g corresponding to the Bash // substitution for replacing all occurrences) but note that in Bash you can avoid the echo and the pipe by doing:

V=$(sed -e 's/XXXXXXX/string1/g;s/YYYYY/string2/g;s/ZZZZZZZ/string3/g' <<< $V)

This feature is actually called a here string.

edited Apr 03 '18 at 19:59

answered Apr 03 '18 at 19:52

ErikMD

13,377
3
35
71

multiple strings replacements on variable

3 Answers3