bash脚本中命令的“源”输出作为多个变量

Question

I have MSI file, which have inside version.ini file which contains multiple variables I need to take out.

For example:

[VERSION]
fileid=...
version=...
option=...

So I have simple command:

cabextract -p -F "version.ini" installer.msi | awk -F '=' '{if ($1 == "fileid" || $1 == "version" || $1 == "language") print tolower($1)"=\""$2"\"";}' > /tmp/tmpoutput
source /tmp/tmpoutput
echo $version

and with that I achieved extracting version.ini from a msi file and extracting from ini file information I want.

But just for curiosity is there any prettier version to take out variables without temporary writing it to a file. It has nosense, but I cannot find another working and simple way to load multiple variables.

If I write command like that:

var=$(cabextract -p -F "version.ini" installer.msi | awk -F '=' '{if ($1 == "fileid" || $1 == "version" || $1 == "language") printtolower($1)"=\""$2"\"";}')

it won't work. I can't put command in $() because then it merge results of awk command. I've tried to put it into array but that also failed.

Any advice?

Answer 1

You can use process substitution here:

source <(cabextract -p -F "version.ini" installer.msi | awk -F '=' '{if ($1 == "fileid" || $1 == "version" || $1 == "language") print tolower($1)"=\""$2"\"";}')

Using this directive output of cabextract will be used for source like a file.