有没有办法清除使用source＆lt;读取的变量？ （grep = ...）？

Question

可以使用.ini从source <(grep = test.ini)文件中读取变量：

$ cat test.ini 
[head]
  shoulders=/hahaha/
  knees=/lololol/
  toes=/kekeke/

$ source <(grep = test.ini)

$ echo $shoulders
/hahaha/

$ echo $knees
/lololol/

$ echo $toes
/kekeke/

我可以手动清除从source <(grep = ...)命令读取的变量，例如

$ unset toes
$ echo toes

但有没有办法自动跟踪从source命令添加哪些变量并将它们全部取消设置？

Answer 1

unset $(awk -F\= '/=/ { printf gensub(" ","","g",$1)" " }' test.ini)

使用awk创建正在设置的变量列表，然后使用此输出运行unset。我们使用gensub来摆脱所设置变量周围的任何空格。

Answer 2

通常情况下，这类事情非常不可靠。但你可以这样做：

set | grep -v ^_ > /tmp/original-vars  # record variables
source <(grep = test.ini)              # read from init file

# now, compare current variables assigned with the original and unset
# those that were not originally assigned while attempting
# to mask out some common internal variables that bash sets but making
# no claims at robustness or safety:
set | grep -v ^_ | diff -u - /tmp/original-vars \
  | awk '/^-/ && NR>1{print $2}' FS=[=-] \
  | while read var; do unset $var; done

YMMV，读者要小心，等等。