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
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。
其他回答
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 ${haystack[@]} | grep -ow "needle" | wc -w)
下面的代码检查给定值是否在数组中,并返回其从零开始的偏移量:
A=("one" "two" "three four")
VALUE="two"
if [[ "$(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。
如何检查一个Bash数组是否包含一个值
假阳性匹配
array=(a1 b1 c1 d1 ee)
[[ ${array[*]} =~ 'a' ]] && echo 'yes' || echo 'no'
# output:
yes
[[ ${array[*]} =~ 'a1' ]] && echo 'yes' || echo 'no'
# output:
yes
[[ ${array[*]} =~ 'e' ]] && echo 'yes' || echo 'no'
# output:
yes
[[ ${array[*]} =~ 'ee' ]] && echo 'yes' || echo 'no'
# output:
yes
精确匹配
为了寻找精确匹配,你的正则表达式模式需要在值的前后添加额外的空格,如(^|[[:space:]])" value "($|[[:space:]])
# Exact match
array=(aa1 bc1 ac1 ed1 aee)
if [[ ${array[*]} =~ (^|[[:space:]])"a"($|[[:space:]]) ]]; then
echo "Yes";
else
echo "No";
fi
# output:
No
if [[ ${array[*]} =~ (^|[[:space:]])"ac1"($|[[:space:]]) ]]; then
echo "Yes";
else
echo "No";
fi
# output:
Yes
find="ac1"
if [[ ${array[*]} =~ (^|[[:space:]])"$find"($|[[:space:]]) ]]; then
echo "Yes";
else
echo "No";
fi
# output:
Yes
有关更多用法示例,示例的来源在这里
我通常编写这类实用程序来操作变量的名称,而不是变量的值,这主要是因为bash不能通过引用传递变量。
下面是一个使用数组名称的版本:
function array_contains # array value
{
[[ -n "$1" && -n "$2" ]] || {
echo "usage: array_contains <array> <value>"
echo "Returns 0 if array contains value, 1 otherwise"
return 2
}
eval 'local values=("${'$1'[@]}")'
local element
for element in "${values[@]}"; do
[[ "$element" == "$2" ]] && return 0
done
return 1
}
这样,问题示例就变成:
array_contains A "one" && echo "contains one"
etc.
结合Beorn Harris和loentar的回答,我们得出了一个更有趣的单行测试:
delim=$'\x1F' # define a control code to be used as more or less reliable delimiter
if [[ "${delim}${array[@]}${delim}" =~ "${delim}a string to test${delim}" ]]; then
echo "contains 'a string to test'"
fi
它不使用额外的函数,不替换测试,并添加了额外的保护,防止使用控制代码作为分隔符偶尔出现错误匹配。
UPD:感谢@ChrisCogdon的注意,这个错误的代码被重写并以https://stackoverflow.com/a/58527681/972463的形式发布。