Bash variable substitution

daYz · 04-14-2006, 03:17 AM

Hi,

My purpose is to compare the output of two commands with diff. I am trying to do this by assigning the two commands to two variables wherafter I use the variables with diff. Please let me know if I should use another approach to achieve my goal, I am new to bash programming.

The problem I have run into is that I have tried to assign a for loop to a variable like this:

Code:

name="for i in unzip; do dpkg -L $i; done"

but I get the next error when I do "$name":

Code:

-bash: for i in unzip ; do dpkg -L  ; done: command not found

Can someone show me what I am doing wrong?

Thanks for your help.

Ben

Hko · 04-14-2006, 04:18 AM

It's not entirely clear to me what it is you want the script to do exactly. Anyways:
You need to prevent that $i is substituted ("expanded" in shell-script parlance) at the first assignment of the string. One way to do that is putting a backslash in fron of the $. Like this:

Code:

name="for i in unzip; do dpkg -L \$i; done"

Also, to have the shell (bash) execute the string, use eval, otherwise bash will treat the entire string (including whitespace) as a single command:

Code:

eval $name

fotoguy · 04-14-2006, 04:20 AM

My programming is not the best, but if you want the value of a variable to contain a executed code try removing the double quotes and use this one instead and see if it will work:

name=`for i in unzip; do dpkg -L $i; done`

daYz · 04-14-2006, 02:16 PM

Hey thanks guys, your explanation helped me real well.

To explain what my goal is please assume there is another command assigned to a variable. For example:

name2="for i in gcc; do dpkg -L \$i; done"

Now I want to compare the output of this command and the previous command we mentioned with diff.

I have tried:

diff $name $name2

diff "$name" "$name2"

but these output errors.

So what command should I use to compare the output of the two commands?

Groeten/Regards,

Ben