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
没有'grep'和循环的一行检查
if ( dlm=$'\x1F' ; IFS="$dlm" ; [[ "$dlm${array[*]}$dlm" == *"$dlm${item}$dlm"* ]] ) ; then
echo "array contains '$item'"
else
echo "array does not contain '$item'"
fi
这种方法既不使用grep这样的外部实用程序,也不使用循环。
这里发生的是:
we use a wildcard substring matcher to find our item in the array that is concatenated into a string; we cut off possible false positives by enclosing our search item between a pair of delimiters; we use a non-printable character as delimiter, to be on the safe side; we achieve our delimiter being used for array concatenation too by temporary replacement of the IFS variable value; we make this IFS value replacement temporary by evaluating our conditional expression in a sub-shell (inside a pair of parentheses)
其他回答
别胡闹了!使您的解决方案简单、干净和可重用。
这些函数负责索引数组和关联数组。可以通过将搜索算法从线性搜索升级为二进制搜索(用于大型数据集)来改进它们。
##
# Determines if a value exists in an array.
###
function hasArrayValue ()
{
local -r needle="{$1:?}"
local -nr haystack="{$2:?}" # Where you pass by reference to get the entire array in one argument.
# Linear search. Upgrade to binary search for large datasets.
for value in "${haystack[@]}"; do
if [[ "$value" == "$needle" ]]; then
return 0
fi
done
return 1
}
##
# Determines if a value exists in an associative array / map.
###
function hasMapValue ()
{
local -r needle="{$1:?}"
local -nr haystack="{$2:?}"
# Linear search. Upgrade to binary search for large datasets.
for value in "${haystack[@]}"; do
if [[ $value == $needle ]]; then
return 0
fi
done
return 1
}
是的,同样的逻辑,但在处理bash时,如果函数的名称可以让您知道迭代的对象(或不迭代的对象),则可能(可能)有用。
考虑到:
array=("something to search for" "a string" "test2000")
elem="a string"
然后简单检查一下:
if c=$'\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)
这是一个小小的贡献:
array=(word "two words" words)
search_string="two"
match=$(echo "${array[@]:0}" | grep -o $search_string)
[[ ! -z $match ]] && echo "found !"
注意:这种方法不区分大小写“两个单词”,但在问题中不需要这样做。
The answer with most votes is very concise and clean, but it can have false positives when a space is part of one of the array elements. This can be overcome when changing IFS and using "${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 $*
结合这里介绍的一些想法,您可以创建一个没有循环的优雅if语句,它可以精确地进行单词匹配。
find="myword"
array=(value1 value2 myword)
if [[ ! -z $(printf '%s\n' "${array[@]}" | grep -w $find) ]]; then
echo "Array contains myword";
fi
这将不会触发一个单词或val,只有整个单词匹配。如果每个数组值包含多个单词,则会中断。