Shell Programming and Scripting

Hello,

I have this problem.

I have script file, e.g.

#!/usr/bin/ksh
echo $MY_DIR
ls -lt $MY_DIR

I want to list the script but with MY_DIR variable expanded. E.g.

MY_DIR=/abc/xyz (in shell MY_DIR is set)

So I want to list the script and see:

#!/usr/bin/ksh
echo /abc/xyz
ls -lt /abc/xyz

I tried both:
a) while read line; do echo $line; done < script.sh
b) awk '{print $0}' script.sh

But both just list the script without substitutin $MY_DIR with its real value.

How to do this?

if you give the absolute path of MY_DIR inside the script and then echo.... it should work fine....

Try:

Code:

while IFS='' read -r line ;do eval echo "$line"; done < script.sh

I tried eval. The problem is that scrip I'm listing is sql and there are things like: "/*" "(c)" which gives:

=> if line of script starts with /* (begin of sql comment) it gives list of directories in root
=> fi line of script starts with * it gives list of directories in current directory
=> for copyright marks like "(c)" error is given and while loop breaks.

can errors be suppressed somehow in eval?

use export command as below.

Code:

export MY_DIR=/abc/xyz  #(in shell MY_DIR is set)

then do the script normally as you write it in before.

How does this work out then:

Code:

while IFS= read -r line ;do eval printf '"%s\n"' "\"$line\""; done < script.sh

Works!

I don't know what this printf does but will soon find out.

Thank You Scrutinizer!