在Bash中,测试数组是否包含某个值的最简单方法是什么?


当前回答

下面是实现这一点的一个小函数。搜索字符串是第一个参数,其余是数组元素:

set +e #otherwise the script will exit on error
containsElement () {
  local e match="$1"
  shift
  for e; do [[ "$e" == "$match" ]] && return 0; done
  return 1
}

该函数的测试运行如下:

$ array=("something to search for" "a string" "test2000")
$ containsElement "a string" "${array[@]}"
$ echo $?
0
$ containsElement "blaha" "${array[@]}"
$ echo $?
1

其他回答

考虑到:

array=("something to search for" "a string" "test2000")
elem="a string"

然后简单检查一下:

if c=$'\x1E' && p="${c}${elem} ${c}" && [[ ! "${array[@]/#/${c}} ${c}" =~ $p ]]; then
  echo "$elem exists in array"
fi

在哪里

c is element separator
p is regex pattern

(单独分配p,而不是直接在[[]]中使用表达式的原因是为了保持bash 4的兼容性)

另一个没有函数的代码:

(for e in "${array[@]}"; do [[ "$e" == "searched_item" ]] && exit 0; done) && echo "found" || echo "not found"

谢谢@Qwerty关于空格的提示!

对应的功能:

find_in_array() {
  local word=$1
  shift
  for e in "$@"; do [[ "$e" == "$word" ]] && return 0; done
  return 1
}

例子:

some_words=( these are some words )
find_in_array word "${some_words[@]}" || echo "expected missing! since words != word"

下面是实现这一点的一个小函数。搜索字符串是第一个参数,其余是数组元素:

set +e #otherwise the script will exit on error
containsElement () {
  local e match="$1"
  shift
  for e; do [[ "$e" == "$match" ]] && return 0; done
  return 1
}

该函数的测试运行如下:

$ array=("something to search for" "a string" "test2000")
$ containsElement "a string" "${array[@]}"
$ echo $?
0
$ containsElement "blaha" "${array[@]}"
$ echo $?
1
for i in "${array[@]}"
do
    if [ "$i" -eq "$yourValue" ] ; then
        echo "Found"
    fi
done

字符串:

for i in "${array[@]}"
do
    if [ "$i" == "$yourValue" ] ; then
        echo "Found"
    fi
done

这是一个小小的贡献:

array=(word "two words" words)  
search_string="two"  
match=$(echo "${array[@]:0}" | grep -o $search_string)  
[[ ! -z $match ]] && echo "found !"  

注意:这种方法不区分大小写“两个单词”,但在问题中不需要这样做。