在bash中的数组运算符中

Question

有没有办法测试数组是否包含指定的元素？

例如,像:

array=(one two three)

if [ "one" in ${array} ]; then
...
fi

Answer 1

for循环可以解决这个问题.

array=(one two three)

for i in "${array[@]}"; do
  if [[ "$i" = "one" ]]; then
    ...
    break
  fi
done

Answer 2

试试这个:

array=(one two three)
if [[ "${array[*]}" =~ "one" ]]; then
  echo "'one' is found"
fi

Answer 3

我的.bashrc文件中有一个“包含”的函数：

contains () 
{ 
    param=$1;
    shift;
    for elem in "$@";
    do
        [[ "$param" = "$elem" ]] && return 0;
    done;
    return 1
}

它可以很好地与数组配合使用：

contains on $array && echo hit || echo miss
  miss
contains one $array && echo hit || echo miss
  hit
contains onex $array && echo hit || echo miss
  miss

但不需要数组：

contains one four two one zero && echo hit || echo miss
  hit