Understanding cut

krishmaths · #1 (**permalink**) 01-22-2009

In our old code, we have the below statements

Code:

LAST_FRIDAY=2009-01-16
echo $LAST_FRIDAY|cut -d '"' -f2

And the output of this is 2009-01-16

The delimiter provided in cut statement is not used in the variable. Also, when we give f1 or f3 or f10 or any other field instead of f2, we still get the output 2009-01-16.

Can someone explain this behaviour please?

zaxxon · #2 (**permalink**) 01-22-2009

As you already said, the delimeter is not used in the variable. Changing the delimeter to "-" might make more sense.

krishmaths · #3 (**permalink**) 01-22-2009

Got clarified now. The intention of the developer was to use

Code:

LAST_FRIDAY="2009-01-16"

and then cut the second field using delimiter '"'

Sorry for the confusion caused. Thanks!

zaxxon · #4 (**permalink**) 01-22-2009

I don't get it - even if he did, when you call the value of the variable with echo, the " are gone anyway using them like in your snippet. So there will be still nothing to cut out.
I mean I don't want to take your happyness away heh

More UNIX and Linux Forum Topics You Might Find Helpful
Thread	Thread Starter	Forum	Replies	Last Post
Understanding ZFS: Compression	iBot	Solaris BigAdmin RSS	0	11-19-2008 05:40 PM
A little help understanding FIFOs?	deckard	Linux	0	11-01-2005 01:46 PM
Understanding traceroutes	Bobby	IP Networking	2	03-14-2005 05:35 PM
need further understanding of init.d	jigarlakhani	UNIX for Advanced & Expert Users	1	09-20-2002 03:11 PM

Thread Tools	Search this Thread
Show Printable Version Email this Page	Search this Thread: Advanced Search
Display Modes	Rate This Thread
Switch to Linear Mode Hybrid Mode Switch to Threaded Mode	Rate This Thread:


	Powered by