是否可以简化此 BASH eval 表达式？

Question

给定以下 Bash shell 脚本摘录：

# The intent is to take the PATH env variable, break it up into its parts, making them
# appear to be command line args (i.e., ``, ``, ...), and then for this example, just
# echo the parts in space delimited form, but we can imagine that we may want to do other
# things with them - this is just sample usage

# Important Requirement/Constraint
# ================================
# Please do not alter the "PATH to , , , ..." portion of the answer or replace the
# Bash ".." range construct with the output of the "seq" command exec'd in a subshell.
# Preferably, the answer should simply consist of the simplification of the last line of
# code - the "eval eval ..." . Also, please don't simplify by collapsing the whole thing
# to just echo "$@" since we may want to work with only some of the parts, and not
# necessarily the first parts, of the path. That is to say that the 1 and $# in the
# {1..$#} range could be replaced with other shell variables or expr., potentially

# Test case
PATH=/usr/local/bin:/usr/bin:/bin

# The code being examined follows

# Set ':' as the input field separator of the path
IFS=: # Or, more appropriately if in a function: local IFS=:

# Parse the PATH environment variable and break it up into its components
set $PATH

# This is the line we want to simplify, if possible, without losing functionality of
# course (see the comment that follows for details)
eval eval echo '\'$(eval 'echo "${1..$#}"')

# Some notes and explanations regarding the functionality and underlying intent of the
# preceding line:
#   - We start by dynamically creating the following construct: ${1..3}
#     since $# is 3 for our example
#   - Use Bash to expand that construct to:   
#     these vars contain the parsed parts of the PATH
#   - Finally, display the three parts of the PATH using echo: echo   
#   - This causes the following text to be sent to STDOUT:
#     /usr/local/bin /usr/bin /bin

那么，是否可以简化前面代码中的 eval eval... 行，但仍然产生所需的输出，对于上面的示例是：

/usr/local/bin /usr/bin /bin

我正在考虑一个解决方案，该解决方案将用 input/output 重定向（可能）或可能导致 reordering/collapsing 的各种命令替换某些 echo 命令需要的 eval 个命令少于示例中使用的命令。

Answer 1

echo "${PATH}" | tr ':' '\n' > stack

count=1

echo "#/bin/sh-" | tr '-' '\n' >> stack2 

while read line 
do
echo "path${count}=${line}" >> stack2
count=$(($count+1))
done < stack

source stack2

现在您已经在其自己的命名变量中获得了路径的每个部分。

Answer 2

贴近原文，你可以做到

IFS=:
set $PATH
echo "$@"

如果你不想改变IFS和PATH，你可以

set $(sed 's/[^=]*=//;s/:/ /g' <<< ${PATH})
echo "$@"

Answer 3

but still produce the desired output,
/usr/local/bin /usr/bin /bin

刚刚：

echo "${PATH//:/ }"

The intent is to take the PATH env variable, break it up into its parts, making them appear to be command line args (i.e., </code>, <code>, ...), and then for this example, just echo the parts in space delimited form, but we can imagine that we may want to do other things with them - this is just sample usage

我不相信未引用的 shell 扩展。

IFS=':' read -ra patharr <<<"$PATH"
set -- "${patharr[@]}"
IFS=' '; printf "%s\n" "${patharr[*]}"

是否可以简化此 BASH eval 表达式？

Is it possible to simplify this BASH eval expression?

bash

shell

scripting