在bash中用其内容替换“源文件"并扩展变量 [英] Replacing 'source file' with its content, and expanding variables, in bash

问题描述

在script.sh中，

In a script.sh,

source a.sh
source b.sh

CMD1
CMD2
CMD3

如何用其内容替换 source * .sh (不执行命令)?我想看看在寻找文件并扩展所有变量之后，bash解释器将执行什么操作.

how can I replace the source *.sh with their content (without executing the commands)? I would like to see what the bash interpreter executes after sourcing the files and expanding all variables.

我知道我可以使用 set -n -v 或运行 bash -n -v script.sh 2> output.sh ，但这不会代替源代码命令(如果a.sh或b.sh包含变量，则更少).

I know I can use set -n -v or run bash -n -v script.sh 2>output.sh, but that would not replace the source commands (and even less if a.sh or b.sh contain variables).

我考虑使用子外壳，但这仍然不能扩展源代码行.我在源代码行之前和之后尝试了 set + n + v 和 set -n -v 的组合，但这仍然行不通.

I thought of using a subshell, but that still doesn't expand the source lines. I tried a combination of set +n +v and set -n -v before and after the source lines, but that still does not work.

我将使用ssh将输出发送到远程计算机.我可以使用<< output.sh 将内容通过管道传递到ssh命令，但是我不能以root用户身份登录到远程计算机，但是我是一个sudoer.因此，我认为我可以创建脚本并将其作为base64编码的字符串发送(使用

I'm going to send that output to a remote machine using ssh. I could use <<output.sh to pipe the content into the ssh command, but I can't log as root onto the remote machine, but I am however a sudoer. Therefore, I thought I could create the script and send it as a base64-encoded string (using that clever trick ) base64 script | ssh remotehost 'base64 -d | sudo bash'

有解决方案吗?还是您有更好的主意?

Is there a solution? Or do you have a better idea?

推荐答案

您可以执行以下操作:

inline.sh:

inline.sh:

#!/usr/bin/env bash
while read line; do
    if [[ "$line" =~ (\.|source)\s+.+ ]]; then
        file="$(echo $line | cut -d' ' -f2)"
        echo "$(cat $file)"
    else
      echo "$line"
    fi
done < "$1"

请注意，这假定存在 source d个文件，并且不处理错误.您还应该处理可能的哈希爆炸.如果 source 文件本身包含 source ，则需要递归应用脚本，例如像(未经测试)这样的东西:

Note this assumes the sourced files exist, and doesn't handle errors. You should also handle possible hashbangs. If the sourced files contain themselves source, you need to apply the script recursively, e.g. something like (not tested):

while egrep -q '^(source|\.)' main.sh; do
    bash inline.sh main.sh > main.sh
done

让我们测试一下

main.sh:

source a.sh
. b.sh

echo cc
echo "$var_a $var_b"

a.sh:

echo aa
var_a="stack" 

b.sh:

echo bb
var_b="overflow"

结果:

bash inline.sh main.sh

echo aa
var_a="stack"
echo bb
var_b="overflow"

echo cc
echo "$var_a $var_b"

bash inline.sh main.sh | bash

aa
bb
cc
stack overflow

顺便说一句，如果您只想查看bash执行的操作，则可以运行

BTW, if you just want to see what bash executes, you can run

bash -x [script]

或远程

ssh user@host -t "bash -x [script]"

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