如何将字符串追加到Bash数组的每个元素？

Question

I have an array in Bash, each element is a string. How can I append another string to each element? In Java, the code is something like:

for (int i=0; i<array.length; i++)
{
    array[i].append("content");
}

Answer 1

You can append a string to every array item even without looping in Bash!

# cf. http://codesnippets.joyent.com/posts/show/1826
array=(a b c d e)
array=( "${array[@]/%/_content}" )
printf '%s\n' "${array[@]}"

Answer 2

As mentioned by hal

  array=( "${array[@]/%/_content}" )

will append the '_content' string to each element.

  array=( "${array[@]/#/prefix_}" )

will prepend 'prefix_' string to each element

Answer 3

Tested, and it works:

array=(a b c d e)
cnt=${#array[@]}
for ((i=0;i<cnt;i++)); do
    array[i]="${array[i]}$i"
    echo "${array[i]}"
done

produces:

a0
b1
c2
d3
e4

EDIT: declaration of the array could be shortened to

array=({a..e})

To help you understand arrays and their syntax in bash the reference is a good start. Also I recommend you bash-hackers explanation.

Answer 4

You pass in the length of the array as the index for the assignment. The length is 1-based and the array is 0-based indexed, so by passing the length in you are telling bash to assign your value to the slot after the last one in the array. To get the length of an array, your use this ${array[@]} syntax.

declare -a array
array[${#array[@]}]=1
array[${#array[@]}]=2
array[${#array[@]}]=3
array[${#array[@]}]=4
array[${#array[@]}]=5
echo ${array[@]}

Produces

1 2 3 4 5