如何检查一个Bash数组是否包含一个值

假阳性匹配

array=(a1 b1 c1 d1 ee)

[[ ${array[*]} =~ 'a' ]] && echo 'yes' || echo 'no'
# output:
yes

[[ ${array[*]} =~ 'a1' ]] && echo 'yes' || echo 'no'
# output:
yes

[[ ${array[*]} =~ 'e' ]] && echo 'yes' || echo 'no'
# output:
yes

[[ ${array[*]} =~ 'ee' ]] && echo 'yes' || echo 'no'
# output:
yes

精确匹配

为了寻找精确匹配，你的正则表达式模式需要在值的前后添加额外的空格，如(^|[[:space:]])" value "($|[[:space:]])

# Exact match

array=(aa1 bc1 ac1 ed1 aee)

if [[ ${array[*]} =~ (^|[[:space:]])"a"($|[[:space:]]) ]]; then
    echo "Yes";
else
    echo "No";
fi
# output:
No

if [[ ${array[*]} =~ (^|[[:space:]])"ac1"($|[[:space:]]) ]]; then
    echo "Yes";
else
    echo "No";
fi
# output:
Yes

find="ac1"
if [[ ${array[*]} =~ (^|[[:space:]])"$find"($|[[:space:]]) ]]; then
    echo "Yes";
else
    echo "No";
fi
# output:
Yes

有关更多用法示例，示例的来源在这里

借鉴Dennis Williamson的答案，下面的解决方案结合了数组、shell-safe引号和正则表达式，以避免需要:遍历循环;使用管道或其他子过程;或者使用非bash实用程序。

declare -a array=('hello, stack' one 'two words' words last)
printf -v array_str -- ',,%q' "${array[@]}"

if [[ "${array_str},," =~ ,,words,, ]]
then
   echo 'Matches'
else
   echo "Doesn't match"
fi

上面的代码通过使用Bash正则表达式来匹配数组内容的字符串化版本。有六个重要的步骤来确保正则表达式匹配不会被数组中的值的巧妙组合所欺骗:

Construct the comparison string by using Bash's built-in printf shell-quoting, %q. Shell-quoting will ensure that special characters become "shell-safe" by being escaped with backslash \. Choose a special character to serve as a value delimiter. The delimiter HAS to be one of the special characters that will become escaped when using %q; that's the only way to guarantee that values within the array can't be constructed in clever ways to fool the regular expression match. I choose comma , because that character is the safest when eval'd or misused in an otherwise unexpected way. Combine all array elements into a single string, using two instances of the special character to serve as delimiter. Using comma as an example, I used ,,%q as the argument to printf. This is important because two instances of the special character can only appear next to each other when they appear as the delimiter; all other instances of the special character will be escaped. Append two trailing instances of the delimiter to the string, to allow matches against the last element of the array. Thus, instead of comparing against ${array_str}, compare against ${array_str},,. If the target string you're searching for is supplied by a user variable, you must escape all instances of the special character with a backslash. Otherwise, the regular expression match becomes vulnerable to being fooled by cleverly-crafted array elements. Perform a Bash regular expression match against the string.

OP自己添加了以下答案，并附上了评论:

在回答和评论的帮助下，经过一些测试，我得出了这个结论:

function contains() {
    local n=$#
    local value=${!n}
    for ((i=1;i < $#;i++)) {
        if [ "${!i}" == "${value}" ]; then
            echo "y"
            return 0
        fi
    }
    echo "n"
    return 1
}

A=("one" "two" "three four")
if [ $(contains "${A[@]}" "one") == "y" ]; then
    echo "contains one"
fi
if [ $(contains "${A[@]}" "three") == "y" ]; then
    echo "contains three"
fi

@ghostdog74关于使用大小写逻辑检查数组包含特定值的回答的一个小补充:

myarray=(one two three)
word=two
case "${myarray[@]}" in  ("$word "*|*" $word "*|*" $word") echo "found" ;; esac

或者打开extglob选项，你可以这样做:

myarray=(one two three)
word=two
shopt -s extglob
case "${myarray[@]}" in ?(*" ")"$word"?(" "*)) echo "found" ;; esac

我们也可以用if语句:

myarray=(one two three)
word=two
if [[ $(printf "_[%s]_" "${myarray[@]}") =~ .*_\[$word\]_.* ]]; then echo "found"; fi

使用grep和printf

在新行上格式化每个数组成员，然后grep这些行。

if printf '%s\n' "${array[@]}" | grep -x -q "search string"; then echo true; else echo false; fi

$ array=("word", "two words")
$ if printf '%s\n' "${array[@]}" | grep -x -q "two words"; then echo true; else echo false; fi
true

注意，这对delimeter和空格没有问题。

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)