删除数组bash中的元素

Question

im trying to delete two elements at same time in my array in bash script my code is

elegidos = (1 2 3 4 5 6 7)
i=0
j=${#elegidos[@]}
delete=($i $j)
while [ $i -le $j ]; do
        #elegidosmenosdos=${elegidos[@]/$i:$j}
        echo ${elegidos[@]/$delete}
        delete=($i $j)
        let "i++"
        let "j--"

done

the output that i have is

1 2 3 4 5 6 7

1 2 3 4 5 6 7

2 3 4 5 6 7

1 3 4 5 6 7

and i need 21 different combinations with five elements using seven numbers output example

1 2 3 4 5

2 3 4 5 6

7 6 5 4 3 . . . . .(21 MORE)

Answer 1

Well, it took a minute to suss out that you were simply wanting to do an end-to-middle delete of two-elements at a time working from the ends of the array, deleting the end nodes at the same time, increment/decrement your counters and repeat until you reached the middle. Why you want to do it this way is a bit of a mystery. You can, of course, simply unset elegidos and unset all values at once. However, if you want to work in from both ends -- that fine too if you have a purpose for it.

You have several problems in your script. In bash, all arrays are zero indexed . Your array has 7-elements, so the valid indexes are 0-6 . Therefore, you j was wrong to begin with. You needed to subtract 1 from the number of elements to get the index for the end-element, eg

i=0
j=$((${#elegidos[@]} - 1))

bash provides a C-style loop that can greatly simplify your task. While you are free to use a while a C-style loop can handle the index increment and decrement seamlessly, eg

for ((; i <= j; i++, j--)); do

note the i <= j . If you have an odd-number of elements in the array, on your last iteration, you will simply be deleting one-value instead of two. To handle that condition you need a simple test within the loop to check whether [ "$i" = "$j" ] (or using the arithmetic comparison (( i == j )) ).

Putting that altogether, you could refine your element removal to empty the array two-elements at a time to something similar to the following:

#!/bin/bash

elegidos=(1 2 3 4 5 6 7)
i=0
j=$((${#elegidos[@]} - 1))
delete=($i $j)

for ((; i <= j; i++, j--)); do
    declare -p elegidos
    unset elegidos[$i]
    [ "$i" != "$j" ] && unset elegidos[$j]
    delete=($i $j)
done

Example Use/Output

$ bash array_del.sh
declare -a elegidos='([0]="1" [1]="2" [2]="3" [3]="4" [4]="5" [5]="6" [6]="7")'
declare -a elegidos='([1]="2" [2]="3" [3]="4" [4]="5" [5]="6")'
declare -a elegidos='([2]="3" [3]="4" [4]="5")'
declare -a elegidos='([3]="4")'

You can see above, that on the first 3-iterations, both end-elements are removed. However, notice on the last removal (there originally being and odd number of elements, only the single-last value is removed on the final iteration.

Look things over and let me know if I captured what you were attempting and whether you have any further questions. If your intent was something else, drop a comment and I'm happy to help further.