从bash数组中删除项目和索引

Question

我正在尝试删除另一个数组中也存在的数组项。例如：

a=(a b c d)
b=(a d)

for ((i=0; i<${#a[@]}; i++)); do
    for j in "${b[@]}" ; do
        [[ $j == ${a[$i]} ]] && unset a[$i]
    done
done

for ((i=0; i<${#a[@]}; i++ )); do
    echo "$i: ${a[$i]}"
done

输出结果为：

0: 
1: b
2: c

如何摆脱虚拟零指数？预期的产出将是：

0: b
1: c

Answer 1

你试过这个吗？

a=(a b c d)
b=(a d)

for i in "${b[@]}"; do
    a=(${a[@]/$i})
done

for ((i=0; i<${#a[@]}; i++ )); do
    echo "$i: ${a[$i]}"
done

Answer 2

尝试类似：

 for i in "${b[@]}"; do ##for every entry in array b
     a=(${a[@]//$i})  ## remove the entry within a that matches exactly what b has
 done