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

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
}

其他回答

下面是实现这一点的一个小函数。搜索字符串是第一个参数,其余是数组元素:

set +e #otherwise the script will exit on error
containsElement () {
  local e match="$1"
  shift
  for e; do [[ "$e" == "$match" ]] && return 0; done
  return 1
}

该函数的测试运行如下:

$ array=("something to search for" "a string" "test2000")
$ containsElement "a string" "${array[@]}"
$ echo $?
0
$ containsElement "blaha" "${array[@]}"
$ echo $?
1

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。

扩展上面来自Sean DiSanti的答案,我认为下面是一个简单而优雅的解决方案,它避免了对数组进行循环,并且不会由于部分匹配而给出假阳性

function is_in_array {
    local ELEMENT="${1}"
    local DELIM=","
    printf "${DELIM}%s${DELIM}" "${@:2}" | grep -q "${DELIM}${ELEMENT}${DELIM}"
}

可以这样称呼:

$ haystack=("needle1" "needle2" "aneedle" "spaced needle")
$ is_in_array "needle" "${haystack[@]}"
$ echo $?
1
$ is_in_array "needle1" "${haystack[@]}"
$ echo $?
0

我有这样的情况,我必须检查一个ID是否包含在另一个脚本/命令生成的ID列表中。 我的工作如下:

# the ID I was looking for
ID=1

# somehow generated list of IDs
LIST=$( <some script that generates lines with IDs> )
# list is curiously concatenated with a single space character
LIST=" $LIST "

# grep for exact match, boundaries are marked as space
# would therefore not reliably work for values containing a space
# return the count with "-c"
ISIN=$(echo $LIST | grep -F " $ID " -c)

# do your check (e. g. 0 for nothing found, everything greater than 0 means found)
if [ ISIN -eq 0 ]; then
    echo "not found"
fi
# etc.

你也可以像这样缩短/压缩它:

if [ $(echo " $( <script call> ) " | grep -F " $ID " -c) -eq 0 ]; then
    echo "not found"
fi

在我的例子中,我正在运行jq来过滤一些JSON的ID列表,然后必须检查我的ID是否在这个列表中,这对我来说是最好的。 它不适用于手动创建的LIST=("1" "2" "4")类型的数组,而是用于换行分隔的脚本输出。


附言:不能评论一个答案,因为我是相对较新的…

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