Assigning outputstring with spaces to variable in Bash script

Question

I want to enter an output of command with a few pipes to var. The code I wrote looks like this:

curl https://www.gentoo.org/downloads/signatures/ | grep 0x | cut -d '>' -f3 | cut -d '<' -f1  |  while read line; do
gpg --recv-keys $line
tempfingerprint= `gpg --fingerprint $line | head -2 | tail -1 | cut -d'=' -f2 | cut -d ' ' -f2-12`
echo $tempfingerprint

cut when I try to echo the results (last code line) I get an error message. I debugged it, and this is the debug log:

336 + head -2
336 + tail -1
336 + cut -d= -f2 
336 + cut -d ' ' -f2-12 
336 + gpg --fingerprint 0xBB572E0E2D182910
36 + tempFingerPrint= 36 + 13EB BDBE DE7A 1277 5DFD B1BA BB57 2E0E 2D18 2910 
./gentoo-stage.sh: line 36: 13EB: command not found

How can I assign the all fingerprint to the variable?

Etan Reisner · Accepted Answer · 2016-03-18 16:00:12Z

2

There's a space after the =:

tempfingerprint= `gpg --fingerprint $line | head -2 | tail -1 | cut -d'=' -f2 | cut -d ' ' -f2-12`
#               ^

That's what causing the error, remove it.

Also, it's not required, but you should prefer "$(...)" over `...`, as it's safer and easier to read:

tempfingerprint="$(gpg --fingerprint $line | head -2 | tail -1 | cut -d'=' -f2 | cut -d ' ' -f2-12)"

And, in general, always quote your variable expansions.

edited Mar 18, 2016 at 16:00

Etan Reisner

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answered Mar 18, 2016 at 15:48

Andrea Corbellini

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