Unix Shell - Passing parmaters to a shell script

Author

Passing parmaters to a shell script

Stu

2004-05-31, 6:06 pm

I have the following shell script:

cat t.sh
==========
spec="SPEC='C:\tmp\00000291.s00'"
sh t1.sh "${spec}"

that calls t1.sh

cat t1.sh
set -x
echo "|$1|"

When I run t.sh I see the following displayed to my screen
t1.sh
+ echo |SPEC='C:\tmp\00000291.s00'|
|SPEC='C: mp0291.s00'|

Can anybody tell why this is appearing ( |SPEC='C: mp0291.s00'| ) and
how to get rid of it. I just want this ( SPEC='C:\tmp\00000291.s00' ) to be
in $1 in t1.sh

Thanks for all that answer

Stephane CHAZELAS

2004-05-31, 6:06 pm

2004-05-28, 07:06(-07), Stu:
> I have the following shell script:
>
> cat t.sh
> ==========
> spec="SPEC='C:\tmp\00000291.s00'"
> sh t1.sh "${spec}"
>
> that calls t1.sh
>
> cat t1.sh
> set -x
> echo "|$1|"

echo can't be use to display arbitrary data as it expands escape
sequences (here \t for tab). Moreover, it's one of the least
portable commands, some of them even accept options (you can't
display "-n" then for instance). It's a command to avoid in
scripts except to display constant strings (litterals).

Use printf instead:

printf '%s\n' "|$1|"

--
Stephane