sh script not finding variables

Question

I am following a tutorial on how to run a reverse ssh tunnel which is found at http://wiki.fabelier.org/index.php?title=Permanent_Reverse_SSH_Tunneling Issue I am having is when I run the tunneling.sh script:

#!/bin/sh
a=`ps -ef | grep 19999 | grep -v grep`
if [ ! "$a" ]; then
    ssh -fN -R 19999:localhost:22 <middle-usename>@<middle-hostname>
fi

I receive this error:

tunnel2.sh: 2: tunnel2.sh: a: not found

EDIT:

I changed shebang to #!/bin/bash

now I get this error:

tunnel2.sh: 2: tunnel2.sh: pi: not found

@PaulTomblin, Sorry I still don't understand. Could you please help if you know the solution — Richard, Dec 04 '12 at 18:30
I don't see any reference to `pi` anywhere in the supplied code. What happens if you just type the commands on the command line instead of running it as a script? — Paul Tomblin, Dec 04 '12 at 18:58
@PaulTomblin from editing several times, I actually had a space line 2. line 2 was actually a= `ps -ef | grep 19999 | grep -v grep` not a=`ps -ef | grep 19999 | grep -v grep`. I don't understand the difference but that fixed it. I guess I will have to do some more reading — Richard, Dec 04 '12 at 19:04
This might help explain the difference: http://stackoverflow.com/questions/2268104/basic-bash-script-variable-declaration-command-not-found/2268117#2268117 — William Pursell, Dec 04 '12 at 22:22
@WilliamPursell Thanks William, That helped. bash scripting is a bit unforgiving. — Richard, Dec 05 '12 at 12:25

score 12 · Accepted Answer · answered Dec 04 '12 at 18:43

12

Don't specify #!/bin/sh in your "shebang" line if you're going to use bash features. If you want bash, ask for bash.

answered Dec 04 '12 at 18:43

Paul Tomblin

thanks Paul, please understand that I am knowledgeable in other languages but I am just starting out in shell scripting. I did try to change the shebang line as suggest but I get a new error as shown above – Richard Dec 04 '12 at 18:46

score 0 · Answer 2 · answered Dec 06 '12 at 04:01

0

Move the ! out of the test [ if you need sh

if ! [ "$a" ]; then

answered Dec 06 '12 at 04:01

Derek Schrock

2 Answers2