一行的解决方案

printf '%s\0' "${myarray[@]}" | grep -F -x -z -- 'myvalue'

解释

printf语句打印数组中的每个元素，以空字符分隔。

grep语句使用以下标志来匹配一个包含myvalue字符串的项(不多不少):

-z/——null-data -行以0字节而不是换行符结束。 -f /——fixed-strings -将pattern解释为固定字符串，而不是正则表达式。 -x/——line-regexp -只选择与整行完全匹配的匹配项。 ——-标记命令行选项的结束，使Grep处理“myvalue”作为一个非选项参数，即使它以破折号开始

为什么我们使用空字节\0而不是换行符\n?数组的元素实际上可能包含换行符。(如果您知道它没有，请随意删除-z grep选项，并将%s\n替换为您的第一个printf参数。)

使用

把这个放进一个if…然后声明:

if printf '%s\0' "${myarray[@]}" | grep -Fxqz -- 'myvalue'; then
    # ...
fi

我在grep表达式中添加了-q标志，这样它就不会打印匹配项;它只会将匹配的存在视为“真”。

更新:感谢presto8指出——line-regexp标志。谢谢Tino，你指出了数组项中可以存在换行符的情况。

我的版本的正则表达式技术，已经建议:

values=(foo bar)
requestedValue=bar

requestedValue=${requestedValue##[[:space:]]}
requestedValue=${requestedValue%%[[:space:]]}
[[ "${values[@]/#/X-}" =~ "X-${requestedValue}" ]] || echo "Unsupported value"

What's happening here is that you're expanding the entire array of supported values into words and prepending a specific string, "X-" in this case, to each of them, and doing the same to the requested value. If this one is indeed contained in the array, then the resulting string will at most match one of the resulting tokens, or none at all in the contrary. In the latter case the || operator triggers and you know you're dealing with an unsupported value. Prior to all of that the requested value is stripped of all leading and trailing whitespace through standard shell string manipulation.

我相信它是干净而优雅的，尽管如果支持的值数组特别大，我不太确定它的性能如何。

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

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。

我通常只使用:

inarray=$(echo ${haystack[@]} | grep -o "needle" | wc -w)

非零值表示找到了匹配。

．.． 实际上，为了解决它不能与needle1和needle2工作的问题，如果你只想要一个精确匹配，没有更多，没有更少，只需在-o后面添加一个w标志，用于整个单词匹配:

inarray=$(echo ${haystack[@]} | grep -ow "needle" | wc -w)

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.