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
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
其他回答
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)
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 [[ $element == "words" ]]
then
echo "Match"
fi
done
fi
这将输出“Checking”和“Match”。使用array=(word "two words" something),它只会输出"Checking"。使用array=(单词“two widgets”什么的)将没有输出。
for i in "${array[@]}"
do
if [ "$i" -eq "$yourValue" ] ; then
echo "Found"
fi
done
字符串:
for i in "${array[@]}"
do
if [ "$i" == "$yourValue" ] ; then
echo "Found"
fi
done
@ghostdog74关于使用大小写逻辑检查数组包含特定值的回答的一个小补充:
myarray=(one two three)
word=two
case "${myarray[@]}" in ("$word "*|*" $word "*|*" $word") echo "found" ;; esac
或者打开extglob选项,你可以这样做:
myarray=(one two three)
word=two
shopt -s extglob
case "${myarray[@]}" in ?(*" ")"$word"?(" "*)) echo "found" ;; esac
我们也可以用if语句:
myarray=(one two three)
word=two
if [[ $(printf "_[%s]_" "${myarray[@]}") =~ .*_\[$word\]_.* ]]; then echo "found"; fi
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"