虽然这里有几个很好的和有用的答案，但我没有找到一个似乎是性能、跨平台和健壮性的正确组合;所以我想分享一下我为我的代码编写的解决方案:

#!/bin/bash

# array_contains "$needle" "${haystack[@]}"
#
# Returns 0 if an item ($1) is contained in an array ($@).
#
# Developer note:
#    The use of a delimiter here leaves something to be desired. The ideal
#    method seems to be to use `grep` with --line-regexp and --null-data, but
#    Mac/BSD grep doesn't support --line-regexp.
function array_contains()
{
    # Extract and remove the needle from $@.
    local needle="$1"
    shift

    # Separates strings in the array for matching. Must be extremely-unlikely
    # to appear in the input array or the needle.
    local delimiter='#!-\8/-!#'

    # Create a string with containing every (delimited) element in the array,
    # and search it for the needle with grep in fixed-string mode.
    if printf "${delimiter}%s${delimiter}" "$@" | \
        grep --fixed-strings --quiet "${delimiter}${needle}${delimiter}"; then
        return 0
    fi

    return 1
}

如果您想做一个快速而复杂的测试，看看是否值得遍历整个数组以获得精确匹配，Bash可以像对待标量一样对待数组。测试标量中的匹配项，如果没有，则跳过循环节省时间。显然你会得到假阳性。

array=(word "two words" words)
if [[ ${array[@]} =~ words ]]
then
    echo "Checking"
    for element in "${array[@]}"
    do
        if [[ $element == "words" ]]
        then
            echo "Match"
        fi
    done
fi

这将输出“Checking”和“Match”。使用array=(word "two words" something)，它只会输出"Checking"。使用array=(单词“two widgets”什么的)将没有输出。

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.

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)

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

The answer with most votes is very concise and clean, but it can have false positives when a space is part of one of the array elements. This can be overcome when changing IFS and using "${array[*]}" instead of "${array[@]}". The method is identical, but it looks less clean. By using "${array[*]}", we print all elements of $array, separated by the first character in IFS. So by choosing a correct IFS, you can overcome this particular issue. In this particular case, we decide to set IFS to an uncommon character $'\001' which stands for Start of Heading (SOH)

$ array=("foo bar" "baz" "qux")
$ IFS=$'\001'
$ [[ "$IFS${array[*]}$IFS" =~ "${IFS}foo${IFS}" ]] && echo yes || echo no
no
$ [[ "$IFS${array[*]}$IFS" =~ "${IFS}foo bar${IFS}" ]] && echo yes || echo no
yes
$ unset IFS

这解决了大多数假阳性问题，但需要一个好的IFS选择。

注意:如果之前设置了IFS，最好保存并重新设置，而不是使用未设置的IFS

相关:

访问bash命令行参数$@ vs $*