views:

62

answers:

2

I tried to run commands using pipes.

Basic:

single="ls -l"
$single

which works as expected

Pipes:

multi="ls -l | grep e"
$multi
ls: |: No such file or directory
ls: grep: No such file or directory
ls: e: No such file or directory

...no surprise

bash < $multi

$multi: ambiguous redirect

next try

bash $multi
/bin/ls: /bin/ls: cannot execute binary file

Only

echo $multi > tmp.sh
bash tmp.sh

worked.

Is there a way to execute more complex commands without creating a script for execution?

+3  A: 

You're demonstrating the difference between the shell and the kernel.

"ls -l" is executable by the system execve() call. You can man execve for details, but that's probably too much detail for you.

"ls -l | grep e" needs shell interpretation to set up the pipe. Without using a shell, the '|' character is just passed into execve() as an argument to ls. This is why you see the "No such file or directory" errors.

Solution:

cmd="ls -l | grep e"
bash -c "$cmd"
bukzor
Thanks, any idea why "bash -c" prints the output in a single line? I would like to have the same output as entered manually at command line.
stacker
@stacker I don't think that's true. I get the same result as at the command line. Perhaps your grep is only returning a single line?
bukzor
@bukzor This happend because I forgot to enclose the variable in double quotes.
stacker
A: 

when you want to run commands with pipes, just run it. Don't ever put the command into a variable and try to run it. Simply execute it

ls -l |grep

If you want to capture the output, use $()

var=$(ls -l |grep .. )

ghostdog74
The commands come from a different server and should be excecuted on the target machine. Thus it can't be hard coded.
stacker
see http://mywiki.wooledge.org/BashFAQ/050
ghostdog74