bash printing and incrementing array values

Question

I'm making a bash script in which I need to print a number while it's incremented like this:

0000
0001
0002
0003
0004

I have made this but is not working:

#!/bin/bash
i=0
pass[0]=0
pass[1]=0
pass[2]=0
pass[3]=0
for i in $(seq 1 9)
    pass[3]="$i"
    echo ${pass[*]}
done

I paste the script on cli and i get this.

$ ~ #!/bin/bash
$ ~ i=0
$ ~ pass[0]=0
$ ~ pass[1]=0
$ ~ pass[2]=0
$ ~ pass[3]=0
$ ~ for i in $(seq 1 9)
>     pass[3]="$i"
bash: error sintáctico cerca del elemento inesperado `pass[3]="$i"'
$ ~     echo ${pass[*]}
0 0 0 0
$ ~ done
bash: error sintáctico cerca del elemento inesperado `done'
$ ~ 

Answer 1

Use this pure bash script:

for ((i=0; i<10; i++)); do
   printf "%04d\n" $i
one

OUTPUT:

0000
0001
0002
0003
0004
0005
0006
0007
0008
0009

Answer 2

#!/bin/bash
i=0
pass[0]=0
pass[1]=0
pass[2]=0
pass[3]=0
for i in $(seq 1 9)
do
    pass[3]="$i"
    echo ${pass[*]}
done

did you forget 'do'

Answer 3

For those of you who like expansions, you can also do:

printf "%s\n" {0001..0009}

or

printf "%.4d\n" {1..9}

No loop!

You can store in an array thus:

$ myarray=( {0001..0009} )
$ printf "%s\n" "${myarray[@]}"
0001
0002
0003
0004
0005
0006
0007
0008
0009
$ echo "${myarray[3]}"
0004

Answer 4

You can do the formatting with seq :

seq -w 0000 0010

(if you don't like the {0000..0010} notation, which is more efficient but doesn't allow parameter substitution.)