UNIX for Dummies Questions & Answers

Hi:

I have a script with the next command line

>set a=(grep -w "something" textfile.txt | grep -w "anotherthing" | cut -c1-11)

In the file textfile.txt there are some file names and their paths so when I use this command line it returns a path, for example /directory, and stores in a.

when I try
>cd $a
it comes with the error "Too many arguments", I know that $a is /directory, but if i do >echo $a it returns the command line, this is why i cant use cd $a.

is there a way to actually store the file path into the variable and not the command line. So when ever do >set | grep a it will acutally display
a /directory

I really need you help.

Thank you

Use:
echo "$a"
to debug the problem.

Sorry i probably dont explained my self, what i am trying to do is
>cd $a
or
>ls $a

but none of this work and i have tried >cd echo $a, and >$a | cd
both for cd and ls... i know that when i execute just $a i get the path that was copied from the textfile.txt, but when i see what the variable contain it contains the command line, eval grep -w "something" textfile.txt | grep -w "anotherthing" | cut -c1-11.

please tell me how to use either cd or ls with the value i get from the execution of the command line (grep -w "something" textfile.txt | grep -w "anotherthing" | cut -c1-11)

Thank you

I think you're probably using csh with that syntax and I don't really know csh real well. Also I'm not sure what you're trying to do. But to get the results of a command line into a variable, try using backquotes. I just tried this with csh...

Code:

%echo 1+2 | bc
3
%set a=`echo 1+2 | bc`
%echo $a
3
%