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
下面是实现这一点的一个小函数。搜索字符串是第一个参数,其余是数组元素:
set +e #otherwise the script will exit on error
containsElement () {
local e match="$1"
shift
for e; do [[ "$e" == "$match" ]] && return 0; done
return 1
}
该函数的测试运行如下:
$ array=("something to search for" "a string" "test2000")
$ containsElement "a string" "${array[@]}"
$ echo $?
0
$ containsElement "blaha" "${array[@]}"
$ echo $?
1
下面的代码检查给定值是否在数组中,并返回其从零开始的偏移量:
A=("one" "two" "three four")
VALUE="two"
if [[ "$(declare -p A)" =~ '['([0-9]+)']="'$VALUE'"' ]];then
echo "Found $VALUE at offset ${BASH_REMATCH[1]}"
else
echo "Couldn't find $VALUE"
fi
匹配是在完整的值上完成的,因此设置VALUE="three"将不匹配。
扩展上面来自Sean DiSanti的答案,我认为下面是一个简单而优雅的解决方案,它避免了对数组进行循环,并且不会由于部分匹配而给出假阳性
function is_in_array {
local ELEMENT="${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 $?
0
我有这样的情况,我必须检查一个ID是否包含在另一个脚本/命令生成的ID列表中。 我的工作如下:
# the ID I was looking for
ID=1
# somehow generated list of IDs
LIST=$( <some script that generates lines with IDs> )
# list is curiously concatenated with a single space character
LIST=" $LIST "
# grep for exact match, boundaries are marked as space
# would therefore not reliably work for values containing a space
# return the count with "-c"
ISIN=$(echo $LIST | grep -F " $ID " -c)
# do your check (e. g. 0 for nothing found, everything greater than 0 means found)
if [ ISIN -eq 0 ]; then
echo "not found"
fi
# etc.
你也可以像这样缩短/压缩它:
if [ $(echo " $( <script call> ) " | grep -F " $ID " -c) -eq 0 ]; then
echo "not found"
fi
在我的例子中,我正在运行jq来过滤一些JSON的ID列表,然后必须检查我的ID是否在这个列表中,这对我来说是最好的。 它不适用于手动创建的LIST=("1" "2" "4")类型的数组,而是用于换行分隔的脚本输出。
附言:不能评论一个答案,因为我是相对较新的…
for i in "${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