[NTLUG:Discuss] bash question: executing a list of commands

[Victor Brilon]

The reason this doesn't work (at least in this example) will become 
evident if you echo $i in your loop. It's being read like this:
ls
-l
ls
-lt
ls
-ltr

And so on.
It's possible to work around this just not easy ;)
The thing I'd recommend to do is something like (assuming "command" is 
the program that generates other commands):

command > tmp_file
bash tmp_file
rm tmp_file

Victor
Lance Simmons wrote:
> Suppose I have a file named foo  with a series of commands I want
> execute.  E.g.,
> 
> 	ls -l
> 	ls -lt
> 	ls -ltr
> 
> How do I execute the lines of foo in order?  I've tried
> 
> 	for i in `cat foo`; do "$i"; done
> 	for i in "`cat foo`"; do "$i"; done
> 
> and several others, but haven't been able to make it work.  
> 
> I've tried using xargs, by editing foo to contain only the lines
> 
> 	-l
> 	-lt
> 	-ltr
> 
> and then doing
> 
> 	xargs -p ls < foo
> 
> but the resulting command line is
> 
> 	ls -l -lt -ltr
> 
> not the desired
> 
> 	ls -l
> 	ls -lt
> 	ls -ltr
> 
> What am I missing?  How do I take an arbitrary list of commands and
> execute them seriatim?
> 
> (As to why I don't just put the commands in a script and execute them
> directly, the answer is that I want to build the list of commands as
> output from another command.  In other words, I want the equivalent of
> the find command's -exec option.)
>