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
一行的解决方案
printf '%s\0' "${myarray[@]}" | grep -F -x -z -- 'myvalue'
解释
printf语句打印数组中的每个元素,以空字符分隔。
grep语句使用以下标志来匹配一个包含myvalue字符串的项(不多不少):
-z/——null-data -行以0字节而不是换行符结束。 -f /——fixed-strings -将pattern解释为固定字符串,而不是正则表达式。 -x/——line-regexp -只选择与整行完全匹配的匹配项。 ——-标记命令行选项的结束,使Grep处理“myvalue”作为一个非选项参数,即使它以破折号开始
为什么我们使用空字节\0而不是换行符\n?数组的元素实际上可能包含换行符。(如果您知道它没有,请随意删除-z grep选项,并将%s\n替换为您的第一个printf参数。)
使用
把这个放进一个if…然后声明:
if printf '%s\0' "${myarray[@]}" | grep -Fxqz -- 'myvalue'; then
# ...
fi
我在grep表达式中添加了-q标志,这样它就不会打印匹配项;它只会将匹配的存在视为“真”。
更新:感谢presto8指出——line-regexp标志。谢谢Tino,你指出了数组项中可以存在换行符的情况。
其他回答
$ myarray=(one two three)
$ case "${myarray[@]}" in *"two"*) echo "found" ;; esac
found
以下是我的看法。
如果可以避免的话,我宁愿不使用bash for循环,因为运行它需要时间。如果有什么东西必须循环,让它是用比shell脚本更低级的语言编写的东西。
function array_contains { # arrayname value
local -A _arr=()
local IFS=
eval _arr=( $(eval printf '[%q]="1"\ ' "\${$1[@]}") )
return $(( 1 - 0${_arr[$2]} ))
}
这是通过创建一个临时关联数组_arr来实现的,它的索引是从输入数组的值派生出来的。(请注意,关联数组在bash 4及以上版本中可用,因此此函数在bash的早期版本中无效。)我们设置$IFS以避免在空格上分词。
该函数不包含显式循环,不过bash内部会遍历输入数组以填充printf。printf格式使用%q来确保输入数据被转义,这样它们就可以安全地用作数组键。
$ a=("one two" three four)
$ array_contains a three && echo BOOYA
BOOYA
$ array_contains a two && echo FAIL
$
注意,这个函数使用的所有东西都是bash内置的,因此没有外部管道拖您的后腿,即使在命令展开中也是如此。
如果你不喜欢使用eval…你可以自由地使用另一种方法。: -)
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 "${some_words[@]}" || echo "expected missing! since words != word"
借鉴Dennis Williamson的答案,下面的解决方案结合了数组、shell-safe引号和正则表达式,以避免需要:遍历循环;使用管道或其他子过程;或者使用非bash实用程序。
declare -a array=('hello, stack' one 'two words' words last)
printf -v array_str -- ',,%q' "${array[@]}"
if [[ "${array_str},," =~ ,,words,, ]]
then
echo 'Matches'
else
echo "Doesn't match"
fi
上面的代码通过使用Bash正则表达式来匹配数组内容的字符串化版本。有六个重要的步骤来确保正则表达式匹配不会被数组中的值的巧妙组合所欺骗:
Construct the comparison string by using Bash's built-in printf shell-quoting, %q. Shell-quoting will ensure that special characters become "shell-safe" by being escaped with backslash \. Choose a special character to serve as a value delimiter. The delimiter HAS to be one of the special characters that will become escaped when using %q; that's the only way to guarantee that values within the array can't be constructed in clever ways to fool the regular expression match. I choose comma , because that character is the safest when eval'd or misused in an otherwise unexpected way. Combine all array elements into a single string, using two instances of the special character to serve as delimiter. Using comma as an example, I used ,,%q as the argument to printf. This is important because two instances of the special character can only appear next to each other when they appear as the delimiter; all other instances of the special character will be escaped. Append two trailing instances of the delimiter to the string, to allow matches against the last element of the array. Thus, instead of comparing against ${array_str}, compare against ${array_str},,. If the target string you're searching for is supplied by a user variable, you must escape all instances of the special character with a backslash. Otherwise, the regular expression match becomes vulnerable to being fooled by cleverly-crafted array elements. Perform a Bash regular expression match against the string.
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