[SOLVED] getting directory name with blank spaces into variable

mia_tech · 03-24-2014, 09:55 AM

I have build a menu into my script that goes something like this:

Code:

jorge@nixlap ~/Music $ ls -d */ | cat -n | sed 's/\///g' 2>/dev/null
     1	adele
     2	Jay-Z-Unplugged
     3	Orishas-A lo cubano
     4	Orishas-Emigrante

now, the user would enter the index number of the artist/album, and that number is passed to a function to get the artist name, all is well but when the folder contins blank spaces I'm running into trouble. I'm only getting the first word of folder. This is what I'm using to get the folder name

Code:

artist=$(ls -d */ | cat -n | sed 's/\///g' | grep $1 | awk '{print $2}')

$1 is the index number pass to function... how could I get the whole dir name when containing blank spaces?

szboardstretcher · 03-24-2014, 09:59 AM

Quotes will fix your problem.

Edit: Sorry,.. as in:

Code:

ls "$artist"

grail · 03-24-2014, 10:02 AM

May I suggest you have a look at the select command.

As for the command you have used, try placing your output in a file and then retry with your awk command and see what goes wrong.

suicidaleggroll · 03-24-2014, 10:18 AM

It's the "print $2" in your awk that's causing the problem. "awk '{print $2}'" prints the second column in the row using whitespace for a delimiter, which means if there is any white space in your artist name, it's just going to grab the first word (awk '{print $3}' would grab the second, $4 the third, etc.). You'll need to re-think how you're grabbing the artist name. Maybe instead of "print the second column" you should "print everything after the first column".

mia_tech · 03-24-2014, 10:22 AM

Quote:

Originally Posted by suicidaleggroll

It's the "print $2" in your awk that's causing the problem. "awk '{print $2}'" prints the second column in the row using whitespace for a delimiter, which means if there is any white space in your artist name, it's just going to grab the first word (awk '{print $3}' would grab the second, $4 the third, etc.). You'll need to re-think how you're grabbing the artist name. Maybe instead of "print the second column" you should "print everything after the first column".

yes, I had already realized that... but should I stick with awk and explore how to do that, or would it be easier by other command?... I only know how to do the basic awk manipulation...

suicidaleggroll · 03-24-2014, 10:29 AM

Quote:

Originally Posted by mia_tech

yes, I had already realized that...

Then why didn't you ask the question you really wanted the answer to, how to print all columns except the first one using awk or another tool?

Or better yet, you could google it. A 2 second google search for "awk print all columns except first" yields all of these suggestions:
http://stackoverflow.com/questions/4...field-with-awk
http://stackoverflow.com/questions/2...th-to-the-last
http://stackoverflow.com/questions/2...-three-columns
http://stackoverflow.com/questions/3...except-1-and-2
http://www.unix.com/shell-programmin...first-one.html
http://superuser.com/questions/21339...ntent-except-1
http://www.cyberciti.biz/faq/unix-li...ields-command/
http://objectmix.com/awk/26528-all-c...xcept-one.html
http://www.theunixcode.com/2013/12/u...h-to-the-last/

I was trying to point you in the right direction so you could find the answer yourself and learn something from this experience. Something something teach a man to fish...

jpollard · 03-24-2014, 11:10 AM

I would suggest switching the application to perl.

Much easier to work with for things like this.

grail · 03-24-2014, 11:24 AM

Did you try looking up my suggestion? select in bash can easily accomplish the task you are after. Admittedly you will get the terminating '/' after each directory name, but its a start
and much much simpler.

mia_tech · 03-25-2014, 09:45 AM

Quote:

Originally Posted by grail

Did you try looking up my suggestion? select in bash can easily accomplish the task you are after. Admittedly you will get the terminating '/' after each directory name, but its a start
and much much simpler.

No, I didn't even use awk.... I used this command which did the work just the same

instead of

Code:

artist=$(ls -d */ | cat -n | sed 's/\///g' | grep $1 | awk '{print $2}')

changed to

Code:

artist=$(ls -d */ | head -"$1" | tail -1 | sed 's/\///g')

szboardstretcher · 03-25-2014, 10:11 AM

Huh. So quotes fixed your issue. Nice.

jpollard · 03-25-2014, 11:40 AM

Quote:

Originally Posted by szboardstretcher

Huh. So quotes fixed your issue. Nice.

That was why I suggested switching to perl - it is much easer to get quoting things right, and the different syntax allows for much more complex things to be done easily.

grail · 03-25-2014, 08:20 PM

Well for others who might view the ticket:

Code:

select name in */ EXIT
do
    [[ "$name" == EXIT ]] && break

    echo "You chose artist ${name%/}"
done