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

有点晚了,但你可以用这个:

#!/bin/bash
# isPicture.sh

FILE=$1
FNAME=$(basename "$FILE") # Filename, without directory
EXT="${FNAME##*.}" # Extension

FORMATS=(jpeg JPEG jpg JPG png PNG gif GIF svg SVG tiff TIFF)

NOEXT=( ${FORMATS[@]/$EXT} ) # Formats without the extension of the input file

# If it is a valid extension, then it should be removed from ${NOEXT},
#+making the lengths inequal.
if ! [ ${#NOEXT[@]} != ${#FORMATS[@]} ]; then
    echo "The extension '"$EXT"' is not a valid image extension."
    exit
fi

其他回答

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

回答完之后,我读到了另一个我特别喜欢的答案,但它有缺陷,遭到了不好评。我受到了启发,这里有两种可行的新方法。

array=("word" "two words") # let's look for "two words"

使用grep和printf:

(printf '%s\n' "${array[@]}" | grep -x -q "two words") && <run_your_if_found_command_here>

使用:

(for e in "${array[@]}"; do [[ "$e" == "two words" ]] && exit 0; done; exit 1) && <run_your_if_found_command_here>

对于not_found结果,添加|| <run_your_if_notfound_command_here>

: NeedleInArgs "$needle" "${haystack[@]}"
: NeedleInArgs "$needle" arg1 arg2 .. argN
NeedleInArgs()
{
local a b;
printf -va '\n%q\n' "$1";
printf -vb '%q\n' "${@:2}";
case $'\n'"$b" in (*"$a"*) return 0;; esac;
return 1;
}

使用:

NeedleInArgs "$needle" "${haystack[@]}" && echo "$needle" found || echo "$needle" not found;

对于bash v3.1及以上版本(printf -v支持) 没有分叉,也没有外部程序 没有循环(除了bash中的内部扩展) 适用于所有可能的值和数组,没有异常,没有什么可担心的

也可以直接使用,比如:

if      NeedleInArgs "$input" value1 value2 value3 value4;
then
        : input from the list;
else
        : input not from list;
fi;

对于从v20.5 b到v3.0的bash, printf缺少-v,因此需要额外的2个fork(但不需要执行,因为printf是bash内置的):

NeedleInArgs()
{
case $'\n'"`printf '%q\n' "${@:2}"`" in
(*"`printf '\n%q\n' "$1"`"*) return 0;;
esac;
return 1;
}

注意,我测试了时间:

check call0:  n: t4.43 u4.41 s0.00 f: t3.65 u3.64 s0.00 l: t4.91 u4.90 s0.00 N: t5.28 u5.27 s0.00 F: t2.38 u2.38 s0.00 L: t5.20 u5.20 s0.00
check call1:  n: t3.41 u3.40 s0.00 f: t2.86 u2.84 s0.01 l: t3.72 u3.69 s0.02 N: t4.01 u4.00 s0.00 F: t1.15 u1.15 s0.00 L: t4.05 u4.05 s0.00
check call2:  n: t3.52 u3.50 s0.01 f: t3.74 u3.73 s0.00 l: t3.82 u3.80 s0.01 N: t2.67 u2.67 s0.00 F: t2.64 u2.64 s0.00 L: t2.68 u2.68 s0.00

Call0和call1是对另一个快速pure-bash变体调用的不同变体 Call2在这里。 N=notfound F=firstmatch L=lastmatch 小写字母为短数组,大写字母为长数组

正如您所看到的,这里的这个变体有一个非常稳定的运行时,所以它不太依赖于匹配位置。运行时主要由数组长度决定。搜索变量的运行时高度依赖于匹配位置。所以在边缘情况下,这个变体可以(快得多)。

但非常重要的是,搜索变量的RAM效率更高,因为这里的这个变量总是将整个数组转换为一个大字符串。

所以如果你的内存很紧,你希望大部分比赛都是早期的,那么就不要在这里使用这个。但是,如果您想要一个可预测的运行时,有很长的数组来匹配(期望延迟或根本不匹配),并且双RAM使用也不是太大的问题,那么这里有一些优势。

定时测试脚本:

in_array()
{
    local needle="$1" arrref="$2[@]" item
    for item in "${!arrref}"; do
        [[ "${item}" == "${needle}" ]] && return 0
    done
    return 1
}

NeedleInArgs()
{
local a b;
printf -va '\n%q\n' "$1";
printf -vb '%q\n' "${@:2}";
case $'\n'"$b" in (*"$a"*) return 0;; esac;
return 1;
}

loop1() { for a in {1..100000}; do "$@"; done }
loop2() { for a in {1..1000}; do "$@"; done }

run()
{
  needle="$5"
  arr=("${@:6}")

  out="$( ( time -p "loop$2" "$3" ) 2>&1 )"

  ret="$?"
  got="${out}"
  syst="${got##*sys }"
  got="${got%"sys $syst"}"
  got="${got%$'\n'}"
  user="${got##*user }"
  got="${got%"user $user"}"
  got="${got%$'\n'}"
  real="${got##*real }"
  got="${got%"real $real"}"
  got="${got%$'\n'}"
  printf ' %s: t%q u%q s%q' "$1" "$real" "$user" "$syst"
  [ -z "$rest" ] && [ "$ret" = "$4" ] && return
  printf 'FAIL! expected %q got %q\n' "$4" "$ret"
  printf 'call:   %q\n' "$3"
  printf 'out:    %q\n' "$out"
  printf 'rest:   %q\n' "$rest"
  printf 'needle: %q\n' "$5"
  printf 'arr:   '; printf ' %q' "${@:6}"; printf '\n'
  exit 1
}

check()
{
  printf 'check %q: ' "$1"
  run n 1 "$1" 1 needle a b c d
  run f 1 "$1" 0 needle needle a b c d
  run l 1 "$1" 0 needle a b c d needle
  run N 2 "$1" 1 needle "${rnd[@]}"
  run F 2 "$1" 0 needle needle "${rnd[@]}"
  run L 2 "$1" 0 needle "${rnd[@]}" needle
  printf '\n'
}

call0() { chk=("${arr[@]}"); in_array "$needle" chk; }
call1() { in_array "$needle" arr; }
call2() { NeedleInArgs "$needle" "${arr[@]}"; }

rnd=()
for a in {1..1000}; do rnd+=("$a"); done

check call0
check call1
check call2

OP自己添加了以下答案,并附上了评论:

在回答和评论的帮助下,经过一些测试,我得出了这个结论:

function contains() {
    local n=$#
    local value=${!n}
    for ((i=1;i < $#;i++)) {
        if [ "${!i}" == "${value}" ]; then
            echo "y"
            return 0
        fi
    }
    echo "n"
    return 1
}

A=("one" "two" "three four")
if [ $(contains "${A[@]}" "one") == "y" ]; then
    echo "contains one"
fi
if [ $(contains "${A[@]}" "three") == "y" ]; then
    echo "contains three"
fi

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"