如何在 POSIX sh 中标记数组? [英] How to mark an array in POSIX sh?

问题描述

在替换 shell 脚本中的外部命令时，我使用了一个数组来摆脱 awk 的 NF.

While replacing external commands in a shell script, I used an array to get rid of awk's NF.

现在，由于我从 bash 转移到 POSIX sh，我无法正确标记数组:

Now, since I moved from bash to POSIX sh, I cannot get the array marked right:

#!/bin/bash
export RANGE="0 1 4 6 8 16 24 46 53"
RANGE=($RANGE)
echo arrayelements: $((${#RANGE[@]}))
LAST=$((${#RANGE[@]}-1))
echo "Last element(replace NF): ${RANGE[$LAST]}"

# ./foo
arrayelements: 9
Last element(replace NF): 53

我使用的是 OpenBSD 的 sh，它的大小与 ksh 完全相同.把上面改成/bin/sh，下面好像不行了:

I'm using OpenBSD's, sh and it has exactly the same size as the ksh. Changing above to /bin/sh, it seems that the following doesn't work:

set -A "$RANGE"
set -- "$RANGE"

如何在/bin/sh 中实现上述脚本?(请注意，如果您使用 --posix 调用 bash，它可以正常工作，这不是我要寻找的.)

How could I realise the above script in /bin/sh? (Note that it works fine if you invoke bash with --posix, that's not what I look for.)

推荐答案

数组不是 POSIX 的一部分sh 规范.

Arrays are not part of the POSIX sh specification.

有多种其他方法可以找到最后一个项目.几种可能性:

There are various other ways to find the last item. A couple of possibilities:

#!/bin/sh
export RANGE="0 1 4 6 8 16 24 46 53"
for LAST_ITEM in $RANGE; do true; done
echo "Last element(replace NF): $LAST_ITEM"

或:

#!/bin/sh
export RANGE="0 1 4 6 8 16 24 46 53"
LAST_ITEM="${RANGE##* }"
echo "Last element(replace NF): $LAST_ITEM"

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