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'contains'函数：

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

它适用于数组：

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

Answer 4

我喜欢用grep：

if echo ${array[@]} | grep -qw one; then
  # "one" is in the array
  ...
fi

（请注意，-q和-w都是grep的非标准选项：-w告诉它只处理整个单词，-q（“安静” ）抑制所有输出。）

Answer 5

array="one two three"
if [ $(echo "$array" | grep one | wc -l) -gt 0 ] ; 
  then echo yes; 
fi

如果这很难看，你可以将它隐藏在一个函数中。

Answer 6

如果你只想检查一个元素是否在数组中，另一种方法

case "${array[@]/one/}" in 
 "${array[@]}" ) echo "not in there";;
 *) echo "found ";;
esac

Answer 7

In_array() {
    local NEEDLE="$1"
    local ELEMENT

    shift

    for ELEMENT; do
        if [ "$ELEMENT" == "$NEEDLE" ]; then
            return 0
        fi
    done

    return 1
}

declare -a ARRAY=( "elem1" "elem2" "elem3" )
if In_array "elem1" "${ARRAY[@]}"; then
...

以上的优雅版本。

Answer 8

in_array() {
    local needle=$1 el
    shift
    for el in "$@"; do
        if [ "$el" = "$needle" ]; then
            return 0
        fi
    done
    return 1
}

if in_array 1 1 2 3; then
    echo true
else
    echo false
fi

# alternatively
a=(1 2 3)
if in_array 1 "${a[@]}"; then
...

Answer 9

OPTIONS=('-q','-Q','-s','-S')

find="$(grep "\-q" <<< "${OPTIONS[@]}")"
if [ "$find" = "${OPTIONS[@]}" ];
then
    echo "arr contains -q"
fi

在bash中的数组运算符中

9 个答案: