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 $*

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
}

这对我来说很管用:

# traditional system call return values-- used in an `if`, this will be true when returning 0. Very Odd.
contains () {
    # odd syntax here for passing array parameters: http://stackoverflow.com/questions/8082947/how-to-pass-an-array-to-a-bash-function
    local list=$1[@]
    local elem=$2

    # echo "list" ${!list}
    # echo "elem" $elem

    for i in "${!list}"
    do
        # echo "Checking to see if" "$i" "is the same as" "${elem}"
        if [ "$i" == "${elem}" ] ; then
            # echo "$i" "was the same as" "${elem}"
            return 0
        fi
    done

    # echo "Could not find element"
    return 1
}

示例调用:

arr=("abc" "xyz" "123")
if contains arr "abcx"; then
    echo "Yes"
else
    echo "No"
fi

containsElement () { for e in "${@:2}"; do [[ "$e" = "$1" ]] && return 0; done; return 1; }

现在正确处理空数组。

a=(b c d)

if printf '%s\0' "${a[@]}" | grep -Fqxz c
then
  echo 'array “a” contains value “c”'
fi

如果你喜欢，你可以使用相同的长选项:

--fixed-strings --quiet --line-regexp --null-data

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。