I have an array list
in a bash script, and a variable var
. I know that $var
appears in ${list[@]}
, but have no easy way of determining its index. I'd like to remove it from list
.
This answer achieves something very close to what I need, except that list
retains an empty element where $var
once was. Note, e.g.:
$ list=(one two three)
$ var="two"
$ list=( "${list[@]/$var}" )
$ echo ${list[@]}
one three
$ echo ${#list[@]}
3
The same thing happens if I use delete=( "$var" )
and replace $var
for $delete
in the third line. Also, doing list=( "${list[@]/$var/}" )
makes no difference either.
(I'll note that, experimenting with the comment to that answer, I managed to match only whole words using list=( "${list[@]/%$var}" )
, omitting the #
.)
I also saw this answer proposing a nice trick to keep track of index and use unset
, but that is unfeasible in my case. Finally, the same issue also appeared here, except that OP was satisfied with the result and probably didn't run into the problem empty elements create for me later on in my script, when I iterate through list
. I tried to negate that problem by using expansion as follows, without any apparent effect:
for item in "${list[@]}"; do
if [ -n ${item:+'x'} ];then
...
fi
done
It's the same when I do [ ${#item} > 0 ]
, and I'm running out of ideas. Suggestions?
EDIT:
I have no understanding of why this happens, but @l0b0's comment made me notice something. Using the above preamble, I get:
$ for item in "${list[@]}"; do echo "Here!"; done
Here!
Here!
Here!
but:
$ for item in ${list[@]}; do echo "Here!"; done
Here!
Here!
I'm not sure I can omit the quotes in my script, though, as items are considerably more complicated there (file names and paths, both containing spaces and odd characters).
To delete an element from the array, we can use the command unset. The command takes in the name of the variable in our case the array name and the index of that element.
pop() function: This method is use to remove elements from the end of an array. shift() function: This method is use to remove elements from the start of an array. splice() function: This method is use to remove elements from the specific index of an array.
Answer: Use the splice() Method You can use the splice() method to remove the item from an array at specific index in JavaScript. The syntax for removing array elements can be given with splice(startIndex, deleteCount) .
You can delete an element from existing array though the whole process isn't very straightforward and may appear like a hack.
#!/bin/bash
list=( "one" "two" "three" "four" "five" )
var1="two"
var2="four"
printf "%s\n" "Before:"
for (( i=0; i<${#list[@]}; i++ )); do
printf "%s = %s\n" "$i" "${list[i]}";
done
for (( i=0; i<${#list[@]}; i++ )); do
if [[ ${list[i]} == $var1 || ${list[i]} == $var2 ]]; then
list=( "${list[@]:0:$i}" "${list[@]:$((i + 1))}" )
i=$((i - 1))
fi
done
printf "\n%s\n" "After:"
for (( i=0; i<${#list[@]}; i++ )); do
printf "%s = %s\n" "$i" "${list[i]}";
done
This script outputs:
Before:
0 = one
1 = two
2 = three
3 = four
4 = five
After:
0 = one
1 = three
2 = five
Key part of the script is:
list=( "${list[@]:0:$i}" "${list[@]:$((i + 1))}" )
Here we re-construct your existing array by specifying the index and length to remove the element from array completely and re-order the indices.
If you want to delete the array element & shift the indices, you can use answer by l0b0 or JS웃.
However, if you don't want to shift the indices, you can use below script-let: (Particularly useful for associative arrays)
$ list=(one two three)
$ delete_me=two
$ for i in ${!list[@]};do
if [ "${list[$i]}" == "$delete_me" ]; then
unset list[$i]
fi
done
$ for i in ${!list[@]};do echo "$i = ${list[$i]}"; done
0 = one
2 = three
If you want to shift the indices to make them continuous, re-construct the array as this:
$ list=("${list[@]}")
$ for i in ${!list[@]};do echo "$i = ${list[$i]}"; done
0 = one
1 = three
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