Bash for loop skip first element of array

Question

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.

What does echo $build_list print?

Daemon Painter
– Daemon Painter

2019-08-21 06:47:19 +00:00
Commented Aug 21, 2019 at 6:47 — Daemon Painter
– Daemon Painter, Commented Aug 21, 2019 at 6:47
Better, what does declare -p build_list output

David C. Rankin
– David C. Rankin

2019-08-21 06:53:42 +00:00
Commented Aug 21, 2019 at 6:53 — David C. Rankin
– David C. Rankin, Commented Aug 21, 2019 at 6:53
It works using build_list=($build_string)

unrue
– unrue

2019-08-21 08:47:37 +00:00
Commented Aug 21, 2019 at 8:47 — unrue
– unrue, Commented Aug 21, 2019 at 8:47

Daemon Painter · Accepted Answer · 2019-08-21 06:52:55Z

7

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.

answered Aug 21, 2019 at 6:52

Daemon Painter

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Comments

prefire · Accepted Answer · 2019-08-21 06:48:26Z

4

Replace the second line with this:

build_list=($build_string)

answered Aug 21, 2019 at 6:48

prefire

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Comments

Zavi · Accepted Answer · 2019-08-21 10:27:09Z

1

build_list=("$build_string")

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

edited Aug 21, 2019 at 10:27

Zavi

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answered Aug 21, 2019 at 10:01

黄轩辕

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