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

没有'grep'和循环的一行检查

if ( dlm=$'\x1F' ; IFS="$dlm" ; [[ "$dlm${array[*]}$dlm" == *"$dlm${item}$dlm"* ]] ) ; then
  echo "array contains '$item'"
else
  echo "array does not contain '$item'"
fi

这种方法既不使用grep这样的外部实用程序,也不使用循环。

这里发生的是:

we use a wildcard substring matcher to find our item in the array that is concatenated into a string; we cut off possible false positives by enclosing our search item between a pair of delimiters; we use a non-printable character as delimiter, to be on the safe side; we achieve our delimiter being used for array concatenation too by temporary replacement of the IFS variable value; we make this IFS value replacement temporary by evaluating our conditional expression in a sub-shell (inside a pair of parentheses)

其他回答

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
example:
$ 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。

扩展上面来自Sean DiSanti的答案,我认为下面是一个简单而优雅的解决方案,它避免了对数组进行循环,并且不会由于部分匹配而给出假阳性

function is_in_array {
    local ELEMENT="${1}"
    local DELIM=","
    printf "${DELIM}%s${DELIM}" "${@:2}" | grep -q "${DELIM}${ELEMENT}${DELIM}"
}

可以这样称呼:

$ haystack=("needle1" "needle2" "aneedle" "spaced needle")
$ is_in_array "needle" "${haystack[@]}"
$ echo $?
1
$ is_in_array "needle1" "${haystack[@]}"
$ echo $?
0

下面的代码检查给定值是否在数组中,并返回其从零开始的偏移量:

A=("one" "two" "three four")
VALUE="two"

if [[ "$(declare -p A)" =~ '['([0-9]+)']="'$VALUE'"' ]];then
  echo "Found $VALUE at offset ${BASH_REMATCH[1]}"
else
  echo "Couldn't find $VALUE"
fi

匹配是在完整的值上完成的,因此设置VALUE="three"将不匹配。

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。

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