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
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 $find) ]]; then
echo "Array contains myword";
fi
这将不会触发一个单词或val,只有整个单词匹配。如果每个数组值包含多个单词,则会中断。
其他回答
虽然这里有几个很好的和有用的答案,但我没有找到一个似乎是性能、跨平台和健壮性的正确组合;所以我想分享一下我为我的代码编写的解决方案:
#!/bin/bash
# array_contains "$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
}
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 == $needle ]]; then
return 0
fi
done
return 1
}
是的,同样的逻辑,但在处理bash时,如果函数的名称可以让您知道迭代的对象(或不迭代的对象),则可能(可能)有用。
另一个没有函数的代码:
(for e in "${array[@]}"; do [[ "$e" == "searched_item" ]] && exit 0; done) && echo "found" || echo "not found"
谢谢@Qwerty关于空格的提示!
对应的功能:
find_in_array() {
local word=$1
shift
for e in "$@"; do [[ "$e" == "$word" ]] && return 0; done
return 1
}
例子:
some_words=( these are some words )
find_in_array word "${some_words[@]}" || echo "expected missing! since words != word"
我通常只使用:
inarray=$(echo ${haystack[@]} | grep -o "needle" | wc -w)
非零值表示找到了匹配。
... 实际上,为了解决它不能与needle1和needle2工作的问题,如果你只想要一个精确匹配,没有更多,没有更少,只需在-o后面添加一个w标志,用于整个单词匹配:
inarray=$(echo ${haystack[@]} | grep -ow "needle" | wc -w)
@ghostdog74关于使用大小写逻辑检查数组包含特定值的回答的一个小补充:
myarray=(one two three)
word=two
case "${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