how can I use a variable as a variable name in this bash script

Question

I am currently trying to make a variable name that would consist of another variable

while [ "$countf" -le 9 ]; do
                    vname=$( echo fcp"$countf" )
                    $vname=$( awk -F, -vs="\$fc$countf" '{for (i=1;i<=NF;i++)if($i~"^"s"$"){print i;exit;}}{print "not found"}' <<< $first_line )
                    countf=$(( countf + 1 ))
            done

although when I go to execute the the script that includes the code, something along the lines of the following is outputted:

fcp1=not: command not found

fcp1 being the content of the vname variable. I've tried several different solutions but have not gotten anything to work yet as of right now, if someone could point out what I am doing wrong though I would really appreciate it, thanks.

why do you need this? Why don't you just call it p1p2 or directly cat1? — alestanis
– alestanis, Commented Nov 19, 2012 at 21:13
Repeat of question stackoverflow.com/questions/13462936/… by same user, just 1 hour ago? — anishsane
– anishsane, Commented Nov 20, 2012 at 5:38

Majid Laissi · Accepted Answer · 2012-11-19 21:17:13Z

2

try this:

eval cat$p2="something"

without the quotes around the variable name.

answered Nov 19, 2012 at 21:17

Majid Laissi

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