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
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
其他回答
如果您想做一个快速而复杂的测试,看看是否值得遍历整个数组以获得精确匹配,Bash可以像对待标量一样对待数组。测试标量中的匹配项,如果没有,则跳过循环节省时间。显然你会得到假阳性。
array=(word "two words" words)
if [[ ${array[@]} =~ words ]]
then
echo "Checking"
for element in "${array[@]}"
do
if [[ $element == "words" ]]
then
echo "Match"
fi
done
fi
这将输出“Checking”和“Match”。使用array=(word "two words" something),它只会输出"Checking"。使用array=(单词“two widgets”什么的)将没有输出。
扩展上面来自Sean DiSanti的答案,我认为下面是一个简单而优雅的解决方案,它避免了对数组进行循环,并且不会由于部分匹配而给出假阳性
function is_in_array {
local ELEMENT="${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 $?
0
保持简单:
Array1=( "item1" "item2" "item3" "item-4" )
var="item3"
count=$(echo ${Array1[@]} | tr ' ' '\n' | awk '$1 == "'"$var"'"{print $0}' | wc -l)
[ $count -eq 0 ] && echo "Not found" || echo "found"
: NeedleInArgs "$needle" "${haystack[@]}"
: NeedleInArgs "$needle" arg1 arg2 .. argN
NeedleInArgs()
{
local a b;
printf -va '\n%q\n' "$1";
printf -vb '%q\n' "${@:2}";
case $'\n'"$b" in (*"$a"*) return 0;; esac;
return 1;
}
使用:
NeedleInArgs "$needle" "${haystack[@]}" && echo "$needle" found || echo "$needle" not found;
对于bash v3.1及以上版本(printf -v支持) 没有分叉,也没有外部程序 没有循环(除了bash中的内部扩展) 适用于所有可能的值和数组,没有异常,没有什么可担心的
也可以直接使用,比如:
if NeedleInArgs "$input" value1 value2 value3 value4;
then
: input from the list;
else
: input not from list;
fi;
对于从v20.5 b到v3.0的bash, printf缺少-v,因此需要额外的2个fork(但不需要执行,因为printf是bash内置的):
NeedleInArgs()
{
case $'\n'"`printf '%q\n' "${@:2}"`" in
(*"`printf '\n%q\n' "$1"`"*) return 0;;
esac;
return 1;
}
注意,我测试了时间:
check call0: n: t4.43 u4.41 s0.00 f: t3.65 u3.64 s0.00 l: t4.91 u4.90 s0.00 N: t5.28 u5.27 s0.00 F: t2.38 u2.38 s0.00 L: t5.20 u5.20 s0.00
check call1: n: t3.41 u3.40 s0.00 f: t2.86 u2.84 s0.01 l: t3.72 u3.69 s0.02 N: t4.01 u4.00 s0.00 F: t1.15 u1.15 s0.00 L: t4.05 u4.05 s0.00
check call2: n: t3.52 u3.50 s0.01 f: t3.74 u3.73 s0.00 l: t3.82 u3.80 s0.01 N: t2.67 u2.67 s0.00 F: t2.64 u2.64 s0.00 L: t2.68 u2.68 s0.00
Call0和call1是对另一个快速pure-bash变体调用的不同变体 Call2在这里。 N=notfound F=firstmatch L=lastmatch 小写字母为短数组,大写字母为长数组
正如您所看到的,这里的这个变体有一个非常稳定的运行时,所以它不太依赖于匹配位置。运行时主要由数组长度决定。搜索变量的运行时高度依赖于匹配位置。所以在边缘情况下,这个变体可以(快得多)。
但非常重要的是,搜索变量的RAM效率更高,因为这里的这个变量总是将整个数组转换为一个大字符串。
所以如果你的内存很紧,你希望大部分比赛都是早期的,那么就不要在这里使用这个。但是,如果您想要一个可预测的运行时,有很长的数组来匹配(期望延迟或根本不匹配),并且双RAM使用也不是太大的问题,那么这里有一些优势。
定时测试脚本:
in_array()
{
local needle="$1" arrref="$2[@]" item
for item in "${!arrref}"; do
[[ "${item}" == "${needle}" ]] && return 0
done
return 1
}
NeedleInArgs()
{
local a b;
printf -va '\n%q\n' "$1";
printf -vb '%q\n' "${@:2}";
case $'\n'"$b" in (*"$a"*) return 0;; esac;
return 1;
}
loop1() { for a in {1..100000}; do "$@"; done }
loop2() { for a in {1..1000}; do "$@"; done }
run()
{
needle="$5"
arr=("${@:6}")
out="$( ( time -p "loop$2" "$3" ) 2>&1 )"
ret="$?"
got="${out}"
syst="${got##*sys }"
got="${got%"sys $syst"}"
got="${got%$'\n'}"
user="${got##*user }"
got="${got%"user $user"}"
got="${got%$'\n'}"
real="${got##*real }"
got="${got%"real $real"}"
got="${got%$'\n'}"
printf ' %s: t%q u%q s%q' "$1" "$real" "$user" "$syst"
[ -z "$rest" ] && [ "$ret" = "$4" ] && return
printf 'FAIL! expected %q got %q\n' "$4" "$ret"
printf 'call: %q\n' "$3"
printf 'out: %q\n' "$out"
printf 'rest: %q\n' "$rest"
printf 'needle: %q\n' "$5"
printf 'arr: '; printf ' %q' "${@:6}"; printf '\n'
exit 1
}
check()
{
printf 'check %q: ' "$1"
run n 1 "$1" 1 needle a b c d
run f 1 "$1" 0 needle needle a b c d
run l 1 "$1" 0 needle a b c d needle
run N 2 "$1" 1 needle "${rnd[@]}"
run F 2 "$1" 0 needle needle "${rnd[@]}"
run L 2 "$1" 0 needle "${rnd[@]}" needle
printf '\n'
}
call0() { chk=("${arr[@]}"); in_array "$needle" chk; }
call1() { in_array "$needle" arr; }
call2() { NeedleInArgs "$needle" "${arr[@]}"; }
rnd=()
for a in {1..1000}; do rnd+=("$a"); done
check call0
check call1
check call2
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.