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
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}${delimiter}"; then
return 0
fi
return 1
}
其他回答
我提出了这个方法,它只能在zsh中工作,但我认为一般方法是不错的。
arr=( "hello world" "find me" "what?" )
if [[ "${arr[@]/#%find me/}" != "${arr[@]}" ]]; then
echo "found!"
else
echo "not found!"
fi
只有当${arr[@]/#pattern/}开始或${arr[@]/%pattern/}结束时,才能从每个元素中取出模式。这两个替换可以在bash中工作,但同时${arr[@]/#%pattern/}只能在zsh中工作。
如果修改后的数组等于原始数组,则不包含该元素。
编辑:
这个在bash中工作:
function contains () {
local arr=(${@:2})
local el=$1
local marr=(${arr[@]/#$el/})
[[ "${#arr[@]}" != "${#marr[@]}" ]]
}
在替换之后,它比较两个数组的长度。如果数组包含该元素,则替换将完全删除该元素,并且计数将不同。
回答完之后,我读到了另一个我特别喜欢的答案,但它有缺陷,遭到了不好评。我受到了启发,这里有两种可行的新方法。
array=("word" "two words") # let's look for "two words"
使用grep和printf:
(printf '%s\n' "${array[@]}" | grep -x -q "two words") && <run_your_if_found_command_here>
使用:
(for e in "${array[@]}"; do [[ "$e" == "two words" ]] && exit 0; done; exit 1) && <run_your_if_found_command_here>
对于not_found结果,添加|| <run_your_if_notfound_command_here>
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)
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)
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