I need to strip the first 0 from all values in an array, e.g. change
array=( 01 02 03 [...] 10 [...] 20 [...] )
to
array=(1 2 3 [...] 10 [...] 20 [...] )
I think I can do this with ${parameter/pattern/string} but I am quite lost with the syntax.
Given that it's an array of numbers, I'd do it arithmetically instead of attempting to perform string replacement:
$ a=( {01..20} )
$ echo "${a[@]}"
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
$ b=()
$ for i in "${a[@]}"; do b+=( $((10#$i)) ); done
$ echo "${b[@]}"
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Here $((10#$i)) would cause the variable i to be evaluated as a base-10 number.
$ array=(01 02 03 10 20)
$ echo "${array[@]#0}"
1 2 3 10 20
shopt -s extglob
b=("${a[@]##+(0)}")
printf "%s\n" "${b[@]}"
extglob is not needed. Though @devnull's solution is probably the sanest.You can use parameter expansion and redefine the array directly:
array=(${array[@]#0})
${string#substring} strips the shortest match of $substring from front of $string.
If you want to strip multiple zeroes, e.g. (001 002) to (1 2), you can use shopt -s extglob and then "${var##*(0)}".
$ a=(01 02 03 10 11)
$ ar=(${a[@]#0})
$ for i in "${ar[@]}"; do echo $i; done
1
2
3
10
11
And with multiple zeroes:
$ a=(01 11 0003)
$ ar=(${a[@]#0})
$ for i in "${ar[@]}"; do echo ${i##*(0)}; done
1
11
003
$ shopt -s extglob
$ for i in "${ar[@]}"; do echo ${i##*(0)}; done
1
11
3
