String replacement in bash - bad substitution error

Question

I newbie in bash scripting but i don't uderstand why it's not work

#!/bin/bash
foo=foobarfoobar
echo ${foo//bar/baz}

bad substitution error on line 3

If you're trying to do a substitution, replacing bar with baz, then this is correct. It works on my machine, printing foobazfoobaz, with bash 3.2.48. — Tom Anderson
– Tom Anderson, Commented Jan 22, 2012 at 11:27
GNU bash, version 4.2.10(1)-release it works in command line but not in the script — Aristarhys
– Aristarhys, Commented Jan 22, 2012 at 11:35
@Aristarhys Can you please do echo $0 and uname -a on the command line and add it to your question. Also, if you can paste the error message from the shell. Do ./script.sh on the command line and show the output in your question — jaypal singh
– jaypal singh, Commented Jan 22, 2012 at 11:37
aristarhys@aristarhys-desktop ~/CADET $ sh test sh test test Linux aristarhys-desktop 3.0.0-12-generic #20-Ubuntu SMP Fri Oct 7 14:50:42 UTC 2011 i686 i686 i386 GNU/Linux test: 5: Bad substitution — Aristarhys
– Aristarhys, Commented Jan 22, 2012 at 11:41

Chris · Accepted Answer · 2012-01-22 11:30:32Z

52

That substitution works fine in Bash 4.2.8 (and looks fine according to the documentation).

My best guess would be that you're not actually using Bash - how are you invoking the script? If you're doing sh script.sh you may well be running it with Dash or something similar (and Dash does indeed give a substitution error on line 3). Try explicitly running it with Bash (bash script.sh).

If it turns out you are actually using Dash, there's some useful information on the differences and how to go back to using Bash (if you want to) here: https://wiki.ubuntu.com/DashAsBinSh

answered Jan 22, 2012 at 11:30

Chris

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4 Comments

Aristarhys Over a year ago

GNU bash, version 4.2.10(1)-release

jaypal singh Over a year ago

If on the command line you did sh scriptname then sh is a link to dash in ubuntu. +1 to @Chris

chernevik Over a year ago

Oof. God bless the Debian maintainers, but this is subtle and confusing. It would be nice if << sh >> calls generating errors somehow included a warning that Dash, not Bash, is in use.

user2023370 Over a year ago

My problem came from an executable bash script called from within a Makefile.

Max Spencer · Accepted Answer · 2012-01-22 11:24:29Z

-2

$ foo=foobarfoobar
$ echo ${foo}/bar/baz
foobarfoobar/bar/baz

Just that you have the braces in the wrong place, but then I am no expert at BASH, so perhaps this isn't the effect you're going for..

answered Jan 22, 2012 at 11:24

Max Spencer

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3 Comments

Aristarhys Over a year ago

${foo//bar/baz} must replace all bar to baz, so echo will be foobazfoobaz tldp.org/LDP/abs/html/string-manipulation.html

jaypal singh Over a year ago

OP is doing global substitution using //. I don't think he wants to append bar/baz to his variable. He is looking to replace occurrences of bar with baz so that his output looks like foobazfoobaz

Aristarhys Over a year ago

yeah in every bash string manipulation article use such code but it does not work for me

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