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

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。

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"

如果你不想重复,这是值得研究的:

#!/bin/bash
myarray=("one" "two" "three");
wanted="two"
if `echo ${myarray[@]/"$wanted"/"WAS_FOUND"} | grep -q "WAS_FOUND" ` ; then
 echo "Value was found"
fi
exit

片段改编自:http://www.thegeekstuff.com/2010/06/bash-array-tutorial/ 我认为这很聪明。

编辑: 你可以这样做:

if `echo ${myarray[@]} | grep -q "$wanted"` ; then
echo "Value was found"
fi

但后者仅在数组包含唯一值时有效。我认为,在143中寻找1只会给出假阳性。

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 $*

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。