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
使用参数展开:
如果参数为空或未设置,则什么都没有 替换,否则词的展开就会被替换。
declare -A myarray
myarray[hello]="world"
for i in hello goodbye 123
do
if [ ${myarray[$i]:+_} ]
then
echo ${!myarray[$i]} ${myarray[$i]}
else
printf "there is no %s\n" $i
fi
done
其他回答
以下是我对这个问题的看法。以下是简短的版本:
function arrayContains() {
local haystack=${!1}
local needle="$2"
printf "%s\n" ${haystack[@]} | grep -q "^$needle$"
}
长一点的版本,我觉得看起来更舒服。
# With added utility function.
function arrayToLines() {
local array=${!1}
printf "%s\n" ${array[@]}
}
function arrayContains() {
local haystack=${!1}
local needle="$2"
arrayToLines haystack[@] | grep -q "^$needle$"
}
例子:
test_arr=("hello" "world")
arrayContains test_arr[@] hello; # True
arrayContains test_arr[@] world; # True
arrayContains test_arr[@] "hello world"; # False
arrayContains test_arr[@] "hell"; # False
arrayContains test_arr[@] ""; # False
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 $?
1
The answer with most votes is very concise and clean, but it can have false positives when a space is part of one of the array elements. This can be overcome when changing IFS and using "${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 $*
回答完之后,我读到了另一个我特别喜欢的答案,但它有缺陷,遭到了不好评。我受到了启发,这里有两种可行的新方法。
array=("word" "two words") # let's look for "two words"
使用grep和printf:
(printf '%s\n' "${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>
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"