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
下面的代码检查给定值是否在数组中,并返回其从零开始的偏移量:
A=("one" "two" "three four")
VALUE="two"
if [[ "$(declare -p A)" =~ '['([0-9]+)']="'$VALUE'"' ]];then
echo "Found $VALUE at offset ${BASH_REMATCH[1]}"
else
echo "Couldn't find $VALUE"
fi
匹配是在完整的值上完成的,因此设置VALUE="three"将不匹配。
其他回答
我看到了一些解决这个问题的方法。
对于使用grep的索引数组
grep ${value} <<< ${array[*]} && true || false
对于使用grep的关联数组键
grep ${value} <<< "${!array[*]}" && true || false
我们可以使用awk,但它可能太过了。
awk --assign "v=${value}" '$v~$0 {print true}' <<<"${!array[*]}
Case语句。
case "${array[*]}" in (*${value}*) true ;; (*) false ;; esac
Bash条件表达式在ksh88风格双方括号:
[[ ${array[@]} =~ ${value} ]] && true || false
注意:顺序很重要,正则表达式在=~ match运算符的右边。
Bash for循环
for ((i=0;i<"${#array[*]}";i++)) ; [[ ${array[i]} = $value ]] && break 0 &> /dev/null || continue; done
注意,在这种特殊情况下,真逻辑是颠倒的,即1=真,0=假。这是因为我们使用break 0强制break内置除true之外的退出代码,除非break n参数小于1,否则总是如此。我们必须要打破循环,我们想要一个布尔退出码除了默认的'true',所以在这种情况下,我们翻转了逻辑。因此,使用具有返回true语义的函数可能更有意义。
借鉴Dennis Williamson的答案,下面的解决方案结合了数组、shell-safe引号和正则表达式,以避免需要:遍历循环;使用管道或其他子过程;或者使用非bash实用程序。
declare -a array=('hello, stack' one 'two words' words last)
printf -v array_str -- ',,%q' "${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 ${array_str},,. If the target string you're searching for is supplied by a user variable, you must escape all instances of the special character with a backslash. Otherwise, the regular expression match becomes vulnerable to being fooled by cleverly-crafted array elements. Perform a Bash regular expression match against the string.
以下是我对这个问题的看法。以下是简短的版本:
function arrayContains() {
local haystack=${!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$"
}
例子:
test_arr=("hello" "world")
arrayContains test_arr[@] hello; # True
arrayContains test_arr[@] world; # True
arrayContains test_arr[@] "hello world"; # False
arrayContains test_arr[@] "hell"; # False
arrayContains test_arr[@] ""; # False
containsElement () { for e in "${@:2}"; do [[ "$e" = "$1" ]] && return 0; done; return 1; }
现在正确处理空数组。
使用grep和printf
在新行上格式化每个数组成员,然后grep这些行。
if printf '%s\n' "${array[@]}" | grep -x -q "search string"; then echo true; else echo false; fi
example:
$ array=("word", "two words")
$ if printf '%s\n' "${array[@]}" | grep -x -q "two words"; then echo true; else echo false; fi
true
注意,这对delimeter和空格没有问题。