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Top Forums Shell Programming and Scripting Reading ls -l output line by line awk the user name and su user to run commands Post 302873595 by Don Cragun on Tuesday 12th of November 2013 12:29:35 PM
Old 11-12-2013
Quote:
Originally Posted by zubairom
Smilie Thanks Made in Germany and Carlos for your input... Now I understand what Carlos meant... Yes echoing the $line works ... That was confusing me as why the why loop goes into the awk subshell...you guys cleared that.

For this script I will use variables as suggested by Don as that is a simpler approach.

regarding the additional checks that you proposed </dev/null work fine but I run into some issues using -f option for su and the 'ticks'... With -f option for su it looks like I can't use -R for chmod(my original script did not have -R for chmod but now I added it) and I get

Code:
-R: 0403-010 A specified flag is not valid for this command.

With the 'ticks' looks like it can't resolve the variable(which is what i thought)

Code:
chmod: /saswork/sastemp/$sas_work_dir: A file or directory in the path name does not exist.

The error message you showed above should not be printed as a result of using the command:
Code:
</dev/null su $sas_user -fc "whoami; chmod 777 /saswork/sastemp/'$sas_work_dir'"

The double quotes around the string whoami; chmod 777 /saswork/sastemp/'$sas_work_dir' make the single quotes regular characters with no special meaning, so $sas_work_dir should be expanded before su is invoked. Assuming as an example that $sas_work_dir expands to some/dir), the shell should invoke su with a final operand that is the string whoami; chmod 777 /saswork/sastemp/'some/dir' and the shell that su invokes would then remove those single quotes.

Please show us the exact command line that you are using to invoke su.
 

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