Run an application in shell script


 
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# 1  
Run an application in shell script

I run the following command using the command line. I need to run the same command in shell script :

Code:
./sas.server start


start is a parameter that passes to start.server to startup services within start.server.

the path to sas.server is /path/Lev1

Here is my code:

Code:
#!/bin/sh
set -v
 
  PATH=/path/Lev1
  start=start
  echo $PATH
  echo $start

.$PATH  ./sas.server $start


Last edited by Don Cragun; 07-09-2015 at 03:18 AM.. Reason: Add CODE and ICODE tags.
# 2  
The variable PATH has a very special meaning to the SHELL, never set it to some random string. It needs to the a colon separated list of the utilities that will be used in your script. Normally, it should be set in your login shell initialization scripts and not messed with much after that.

With what you currently have, you are trying to execute a (probably non-existent) directory with the command:
Code:
./$PATH

which expands to the command:
Code:
./path/Lev1

invoked with the arguments ./sas.server and start.

I would guess that you want something more like:
Code:
#!/bin/sh
set -v
 
  SAS_PATH=/path/Lev1
  start=start
  echo "$SAS_PATH"
  echo "$start"

cd "$SAS_PATH"
./sas.server "$start"

This User Gave Thanks to Don Cragun For This Post:
# 3  
Why don't you run it giving the entire path of the command in the script:
Code:
/path/Lev1/sas.server start

?
And, I guess "start" is a string constant passed as a parameter; don't expand it with a $ sign.
This User Gave Thanks to RudiC For This Post:
# 4  
Thanks - You are great
 

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