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
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
其他回答
如果你需要性能,你不希望每次搜索时都要遍历整个数组。
在这种情况下,您可以创建一个表示该数组索引的关联数组(哈希表或字典)。也就是说,它将每个数组元素映射到它在数组中的索引:
make_index () {
local index_name=$1
shift
local -a value_array=("$@")
local i
# -A means associative array, -g means create a global variable:
declare -g -A ${index_name}
for i in "${!value_array[@]}"; do
eval ${index_name}["${value_array[$i]}"]=$i
done
}
然后你可以这样使用它:
myarray=('a a' 'b b' 'c c')
make_index myarray_index "${myarray[@]}"
并像这样测试成员:
member="b b"
# the "|| echo NOT FOUND" below is needed if you're using "set -e"
test "${myarray_index[$member]}" && echo FOUND || echo NOT FOUND
或者:
if [ "${myarray_index[$member]}" ]; then
echo FOUND
fi
请注意,即使在测试值或数组值中存在空格,该解决方案也能正确执行。
作为奖励,您还可以通过以下方式获得数组中值的索引:
echo "<< ${myarray_index[$member]} >> is the index of $member"
OP自己添加了以下答案,并附上了评论:
在回答和评论的帮助下,经过一些测试,我得出了这个结论:
function contains() {
local n=$#
local value=${!n}
for ((i=1;i < $#;i++)) {
if [ "${!i}" == "${value}" ]; then
echo "y"
return 0
fi
}
echo "n"
return 1
}
A=("one" "two" "three four")
if [ $(contains "${A[@]}" "one") == "y" ]; then
echo "contains one"
fi
if [ $(contains "${A[@]}" "three") == "y" ]; then
echo "contains three"
fi
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)
The answer with most votes is very concise and clean, but it can have false positives when a space is part of one of the array elements. This can be overcome when changing IFS and using "${array[*]}" instead of "${array[@]}". The method is identical, but it looks less clean. By using "${array[*]}", we print all elements of $array, separated by the first character in IFS. So by choosing a correct IFS, you can overcome this particular issue. In this particular case, we decide to set IFS to an uncommon character $'\001' which stands for Start of Heading (SOH)
$ array=("foo bar" "baz" "qux")
$ IFS=$'\001'
$ [[ "$IFS${array[*]}$IFS" =~ "${IFS}foo${IFS}" ]] && echo yes || echo no
no
$ [[ "$IFS${array[*]}$IFS" =~ "${IFS}foo bar${IFS}" ]] && echo yes || echo no
yes
$ unset IFS
这解决了大多数假阳性问题,但需要一个好的IFS选择。
注意:如果之前设置了IFS,最好保存并重新设置,而不是使用未设置的IFS
相关:
访问bash命令行参数$@ vs $*
有点晚了,但你可以用这个:
#!/bin/bash
# isPicture.sh
FILE=$1
FNAME=$(basename "$FILE") # Filename, without directory
EXT="${FNAME##*.}" # Extension
FORMATS=(jpeg JPEG jpg JPG png PNG gif GIF svg SVG tiff TIFF)
NOEXT=( ${FORMATS[@]/$EXT} ) # Formats without the extension of the input file
# If it is a valid extension, then it should be removed from ${NOEXT},
#+making the lengths inequal.
if ! [ ${#NOEXT[@]} != ${#FORMATS[@]} ]; then
echo "The extension '"$EXT"' is not a valid image extension."
exit
fi