Comments on: Using cut – Shellscript string manipulation

By: supertone44

supertone44 — Wed, 08 Jun 2011 16:58:56 +0000

First off, great tutorial and examples. Found exactly what I was needing.

Can anyone explain to me what the difference is between the ` and ‘ characters. For example, in comment 3109 by thelinuxblog.com they suggested:

NAME=`echo $VAR | cut -d \. -f 1`

I previously had tried wrapping it with the single quote character, which just treated the statement as a string. Can anyone shed some light on the significance of the ` character?

Thanks!

By: challaa

challaa — Wed, 01 Jun 2011 07:01:59 +0000

super example for cut command
thanks
challaa

By: Bobo

Bobo — Sat, 28 May 2011 17:11:22 +0000

Brilliant. You made it quick, simple, and easy.
If only all tutorials could be this way. ;)
Thanks!

By: TheLinuxBlog.com

TheLinuxBlog.com — Sun, 12 Dec 2010 15:46:15 +0000

@yaker agreed, I’d like to see the last option, the first is always easy since it should be “1″. To get the last field, you’d have to count the number of fields and specify that, or reverse the string and use cut to get the first (which was last) again.

like: cat test.txt | rev | cut -d , -f 1

By: yaker

yaker — Sun, 12 Dec 2010 14:29:20 +0000

it will be more comfortable if the “-f” param can accept fields such as “first/last”

By: TheLinuxBlog.com

TheLinuxBlog.com — Fri, 29 Jan 2010 16:22:40 +0000

Hey Maga, assuming the filename in VAR doesn’t have any periods in it you can use:

NAME=`echo $VAR | cut -d \. -f 1`

Here is another way subtracting 5 characters (extension + blankspace) from the $VAR and using the character cut:
echo $VAR | cut -c -$((`echo $VAR| wc -c` – 5))

However, there are probably better ways of doing it.

By: Maga

Maga — Fri, 29 Jan 2010 14:57:05 +0000

oops, I forgot to say that I want to store the output in a variable, which was my question….

so I’ve tried:

name=echo “${files[$i]}”| cut -d ‘.’ -f 1

and I don’t get anything stored in the variable….

By: Maga

Maga — Fri, 29 Jan 2010 14:54:25 +0000

Awesome tutorial on cut!!! Now it’s part of my bookmarks… :)

I still have one question… I have the name of a file stored in a variable and want to get only the name, not the extension…
I’ve tried
echo ${files[$i]}| cut -d ‘.’ -f 1 and it doesn’t work…

could anyone tell me what am missing here?

thx!!

By: Roger

Roger — Mon, 05 Oct 2009 21:06:27 +0000

Please ignore my previous post. Should have read the manual before posting !

In bash, the default output delimiter is the same as the input delimiter.

–output-delimiter=STRING
use STRING as the output delimiter the default is to use the input delimiter

By: Roger

Roger — Mon, 05 Oct 2009 20:48:03 +0000

In bash why can’t I split the expression parfile=/tmp/abc.par into its 2 components where the delimiter is an = sign:

#echo parfile=/tmp/abc.par | cut -f 1-2 -d “=”
parfile=/tmp/abc.par

I tried various versions of it (with the -e to the echo, field list specified as 1,2 as well as 1-2, space after the f, no space after the f).

Nothing is working. Can anyone shed some light?

Thanks