Shell Programming and Scripting

Hi

i have a small script like this

Code:

  $ cat test.sh
#!/usr/bin/sh
name="ram"
echo ${1}
set 1 2 3 4 5 6 7 8 9 0 123 o870
echo $9
echo ${12}

when i am trying to execute like below i am not getting the output

Code:

jena samp_perl $ sh test.sh
test.sh: This:  not found

but when i executing like below i am able to get the output

Code:

jena samp_perl $ ./test.sh 1
1
9
o870

so may i know wt is the difference between two types of executions and why i am not able to get the output when i tried like this sh test.sh...


please help me in this.. thanks in advance

Moderator's Comments:

Please use next time code tags for your code and data

you have no value in $1 in the first invocation, thus ${1} fails and your script exits.

It works for me.

Even if ${1} in echo ${1} is not defined, the statement will result to echo and should output an empty line.

Can you paste the output of sh -x test.sh

--ahamed

What happens when you replace echo ${1} with echo "${1}" ?
-also-
What does ./test.sh produce?

I think you have two scripts called test.sh . One is in the current working directory (./test.sh) and the other one is somewhere which can be found though $PATH (sh test.sh).
To find the duff one:

Code:

whence test.sh
# Or if your Shell doesn't have the "whence" command:
type test.sh

Interestingly if your current directory is not in $PATH the "sh test.sh" won't find the test.sh in your current directory.

@Methyl, but according to the POSIX specification:
sh

Perhaps if it is a non-compliant shell ?

On any system I've come across you would only definitely execute a script in the current directory with "sh scriptname" or just "scriptname" if you had a "." at the start of $PATH.
We routinely remove the "." from $PATH because it is a security risk.

Have you come across a unix where "sh scriptname" works for the current directory when there is no "." in $PATH ?

Ps. Forget to suggest checking for an alias.

Code:

alias | grep "test.sh"