CODEWITHSUNDEEP
bash
unrue
unrue

Reputation: 197

Bash for loop skip first element of array

In bash, I have a string, I want to convert in array and use it in a for loop but skipping the first element. The code below does not works:

build_string="100 99 98"
build_list=("$build_string")

echo $build_list
for i in "${build_list[@]:1}"
   do  echo "i: " $i
done

The for loop does not print anything. Could you help me? Thanks.

Upvotes: 5

Views: 13899

Answers (3)

黄轩辕
黄轩辕

Reputation: 11

build_list=("$build_string")

build_list array only have one item, so ${build_list[@]:1 is empty.

Upvotes: 1

Daemon Painter
Daemon Painter

Reputation: 3470

I believe you are not converting the array properly (or at all). Please see this snippet:

build_string="100 99 98"
#build_list=("$build_string")  <-- this is not converting into array, following line is.
IFS=' ' read -r -a build_list <<< "$build_string"

echo $build_list
for i in "${build_list[@]:1}"
   do  echo "i: " $i
done

sleep 2

now the output is:

100
i:  99
i:  98

Which sounds reasonable. The 100 is printed when you ask echo $build_string. Reference: split string into array in bash

As pointed out by prefire, the double quotes on the second line are preventing array conversion. This snippet also works:

build_string="100 99 98"
build_list=($build_string)

echo $build_list
for i in "${build_list[@]:1}"
   do  echo "i: " $i
done

sleep 2

Note: I've added a sleep 2 at the end, so I can see what is being printed.

Upvotes: 7

prefire
prefire

Reputation: 76

Replace the second line with this:

build_list=($build_string)

Upvotes: 4

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