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

containsElement () { for e in "${@:2}"; do [[ "$e" = "$1" ]] && return 0; done; return 1; }

现在正确处理空数组。

其他回答

这种方法的优点是不需要遍历所有元素(至少不是显式地)。但是由于array.c中的array_to_string_internal()仍然循环遍历数组元素并将它们连接到一个字符串中,因此它可能并不比所提出的循环解决方案更有效,但它更具可读性。

if [[ " ${array[*]} " =~ " ${value} " ]]; then
    # whatever you want to do when array contains value
fi

if [[ ! " ${array[*]} " =~ " ${value} " ]]; then
    # whatever you want to do when array doesn't contain value
fi

请注意,如果您正在搜索的值是带有空格的数组元素中的某个单词,则会给出假阳性。例如

array=("Jack Brown")
value="Jack"

正则表达式将“Jack”视为在数组中,即使它不在数组中。所以你必须改变IFS和正则表达式上的分隔符如果你仍然想使用这个解决方案,就像这样

IFS="|"
array=("Jack Brown${IFS}Jack Smith")
value="Jack"

if [[ "${IFS}${array[*]}${IFS}" =~ "${IFS}${value}${IFS}" ]]; then
    echo "true"
else
    echo "false"
fi

unset IFS # or set back to original IFS if previously set

这将打印“false”。

显然,这也可以用作测试语句,允许将其表示为一行程序

[[ " ${array[*]} " =~ " ${value} " ]] && echo "true" || echo "false"

结合Beorn Harris和loentar的回答,我们得出了一个更有趣的单行测试:

delim=$'\x1F' # define a control code to be used as more or less reliable delimiter
if [[ "${delim}${array[@]}${delim}" =~ "${delim}a string to test${delim}" ]]; then
    echo "contains 'a string to test'"
fi

它不使用额外的函数,不替换测试,并添加了额外的保护,防止使用控制代码作为分隔符偶尔出现错误匹配。


UPD:感谢@ChrisCogdon的注意,这个错误的代码被重写并以https://stackoverflow.com/a/58527681/972463的形式发布。

这对我来说很管用:

# traditional system call return values-- used in an `if`, this will be true when returning 0. Very Odd.
contains () {
    # odd syntax here for passing array parameters: http://stackoverflow.com/questions/8082947/how-to-pass-an-array-to-a-bash-function
    local list=$1[@]
    local elem=$2

    # echo "list" ${!list}
    # echo "elem" $elem

    for i in "${!list}"
    do
        # echo "Checking to see if" "$i" "is the same as" "${elem}"
        if [ "$i" == "${elem}" ] ; then
            # echo "$i" "was the same as" "${elem}"
            return 0
        fi
    done

    # echo "Could not find element"
    return 1
}

示例调用:

arr=("abc" "xyz" "123")
if contains arr "abcx"; then
    echo "Yes"
else
    echo "No"
fi
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

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 $*