[SOLVED] Bash gurus: Using 'cut' to assign multiple shell variables?

forbin · 08-08-2010, 10:06 AM

Suppose I have a variable containing a delimited string as follows:

MYSTRING="field1:field2:field3:field4"

Then suppose I want to assign the value of field1 within the string to a new variable. I would often do it like this...

MYVAR1=`echo $MYSTRING|cut -d":" -f1`

That works fine, but things get clumsy when I want to assign all of the fields within the string to their own variables, like this...

MYVAR1=`echo $MYSTRING|cut -d":" -f1`
MYVAR2=`echo $MYSTRING|cut -d":" -f2`
MYVAR3=`echo $MYSTRING|cut -d":" -f3`
MYVAR4=`echo $MYSTRING|cut -d":" -f4`

Is there an elegant way to assign all of the fields within MYSTRING to separate variables with a single command?

forbin · 08-08-2010, 10:13 AM

Better yet, suppose I just want fields 2 and 4? I'm aware that I can get just those fields with the cut command, as follows...

echo $MYSTRING|cut -d":" -f2,4

But how do I assign fields 2 and 4 to separate variables? That's the answer I'm really looking for.

grail · 08-08-2010, 10:17 AM

How about short sed:

Code:

MYVAR=$(echo $MYSTRING | sed 's/:/ /g')

Then simply reference each based on array location: eg. ${MYVAR[0]}

forbin · 08-08-2010, 10:27 AM

Well, dang. Once again I posted too soon. Found my answer. Sorry for the trouble.

forbin · 08-08-2010, 10:28 AM

Grail, that works, too. Maybe even better than the other solution I found. Thanks!

grail · 08-08-2010, 10:29 AM

No probs

Please mark as SOLVED if you have your solution.

druuna · 08-08-2010, 10:36 AM

Quote:

Originally Posted by grail

How about short sed:

Code:

MYVAR=$(echo $MYSTRING | sed 's/:/ /g')

Then simply reference each based on array location: eg. ${MYVAR[0]}

Shouldn't that be: MYVAR=($(echo $MYSTRING | sed 's/:/ /g'))

Your code will just fill MYVAR (or MYVAR[0]) with the echoed input:

Code:

$ MYVAR=$(echo $MYSTRING | sed 's/:/ /g')
$ echo ${MYVAR}
field1 field2 field3 field4
$ echo ${MYVAR[0]}
field1 field2 field3 field4
$ echo ${MYVAR[1]}

$

Maybe bash version depended (tested with bash 3.1.1, no access to a 4 version atm).

Just curious.

grail · 08-08-2010, 11:01 AM

@druuna ... yes you are correct .. my bad .. I had actually thrown mine into a loop

and forgot to then say an array needs the extra brackets.

good catch

druuna · 08-08-2010, 11:05 AM

Which does make me wonder:

Quote:

Originally Posted by forbin

Grail, that works, too. Maybe even better than the other solution I found. Thanks!

grail · 08-08-2010, 11:15 AM

Yeah I get the feeling the OP may have had it in a loop