shell script assigning value of array to variables

Question

I want to assign the value in the filearray to the variable $filename0,$filename1...etc

how can I set automatically increment filename and assign correct value result should be like this $filename0=filearray[0] $filename1=filearray[1]

Right now, this is my code

filearray=('tmp1.txt' 'tmp2.txt' 'tmp3.txt' 'tmp4.txt' 'tmp5.txt')
let i=0
while (( i <= 4 ));
do
        filename=${filearray[i]}
        i=$(( i+1 ))
done

Sorry for the late update:
now, when there is an extension in the filename it will prompt
"syntax error: invalid arithmetic operator (error token is ".txt")"
but is ok when I echo $(filename1)

count=0
for i in ${filearray[@]};
do
        count=$(( count +1 ))
        declare filename$count=$i
done
y=0
while [ $y < $count ]
do
        y=$(( y + 1 ))
        echo $((filename$y))
done

Possible duplicate of Dynamic variable names in Bash

kvantour
– kvantour

2018-07-24 07:45:11 +00:00
Commented Jul 24, 2018 at 7:45 — kvantour
– kvantour, Commented Jul 24, 2018 at 7:45

cdarke · Accepted Answer · 2018-07-24 07:40:55Z

1

It seems you want to create variable names dynamically. While that is possible* there is almost always a better way to do it, using an associative array for example:

filearray=('tmp1.txt' 'tmp2.txt' 'tmp3.txt' 'tmp4.txt' 'tmp5.txt')

declare -A fileassoc

for ((i=0; i < ${#filearray[@]}; i++))
do
    fileassoc["filename$i"]=${filearray[i]}
done

# To illustrate:
for key in "${!fileassoc[@]}"
do
    echo "$key: ${fileassoc[$key]}"
done

# or
for ((i=0; i < ${#filearray[@]}; i++))
do
    key="filename$i"
    echo "$key: ${fileassoc[$key]}"
done

Gives:

filename4: tmp5.txt
filename1: tmp2.txt
filename0: tmp1.txt
filename3: tmp4.txt
filename2: tmp3.txt

(associative arrays are not ordered, they don't have to be)

*using a dangerous command called eval - there be dragons, don't go there, even with a magic ring

edited Jul 24, 2018 at 7:40

answered Jul 24, 2018 at 7:34

cdarke

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2 Comments

Jacky Wong Over a year ago

that's almost what I need!, is there anyway that I can direct use $((filename$i))

cdarke Over a year ago

Yes, but it is insecure, dangerous and bad practice. I would be doing a disservice if I showed you how. What is wrong with the method I show?

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