[SOLVED] import variable from file into current bash session

vincix · 01-10-2018, 08:52 AM

Hi,

Is there a way I could run a line which assigns a variable in a script, so that I could use it within the current bash session? In other words, I would like to use the source command only for a certain line in the file. How can I do that?

pan64 · 01-10-2018, 09:26 AM

Code:

line=$(some process to get the line you need)
eval $line

probably, but this is not the best way. And works only if the line has valid syntax. Otherwise you can somehow grep out the value you need and:

Code:

VAR=$(grep pattern file)

will do the assignment.

vincix · 01-10-2018, 09:50 AM

Right, my problem was that I kept trying to execute the line itself, the line which assigns the variable. So what I'm basically doing is var=$(var=...whatever) or some other var name, but that's irrelevant. Thanks.

vincix · 01-10-2018, 09:57 AM

No, that's not right. I got a little bit ahead of myself. A particular example:

Code:

awk 'NR==3 { print }' script
my_ip=$(ip address | grep inet | grep eth0 | awk '{ print $2 }' | cut -d "/" -f 1)

If I try var=$(awk 'NR==3 { print }'), I'd expect the variable to be assigned the output of the awk command, but that's not the case. When I echo $var, it says:

Code:

my_ip=$(ip address | grep inet | grep eth0 | awk '{ print $2 }' | cut -d "/" -f 1)

Which is normal, of course, because the output of that line is another command in itself. If I try var=$($(awk...)), then I get var=$(ip: command not found
So it doesn't seem to be working after all.

The first solution doesn't work either:

Code:

echo $line
my_ip=$(ip address | grep inet | grep eth0 | awk '{ print $2 }' | cut -d "/" -f 1)
eval $line
[no output]

pan64 · 01-11-2018, 01:05 AM

sorry, I don't understand what you posted.

Code:

ip address | grep inet | grep eth0 | awk '{ print $2 }' | cut -d "/" -f 1

returns an IP address, or nothing. so my_ip will be set to that IP (of your eth0).
when you set mp_ip, eval $line has no any meaning, because the variable line was not set at all.
by the way:

Code:

my_ip=$(ip address | awk -F"[ /]*" '/inet.*eth0/ { print $3 }')

looks better a bit (at least for me)

vincix · 01-11-2018, 02:47 AM

Sorry, I posted in a hurry and wasn't paying attention to some basic things.
You are obviously correct.

And yes, I agree with the fact that your solution is more elegant, especially because you're using only one tool instead of three.
I do have a question, though. How does the * work within the [ /]* regex? If I don't include it, then, in order to get the IP, I have to print $6. If I do, then, ofc, it works as you said it works.
Does it mean any number (including 0) of spaces or slashes? And the reason for it is that the indentation has several spaces? Is there any difference in this case between * and +? I know + means 1 or infinite, but in this case with * I'm guessing you couldn't have zero spaces or slashes as a delimiter. So are they equivalent when it comes to awk delimiters?

Another question:
Is inet.*eth0 a greedy regex? If "eth0" showed up twice, then it would match the string up to the last occurrence, right?

pan64 · 01-11-2018, 03:41 AM

about [ /]*: you are right, + can be used too, zero is meaningless in this case, because -F specifies a delimiter, which cannot be empty (in this case, but there are cases where it is accepted).
with * and/or + subsequent spaces will count as one single delimiter, without that you need to use $6 - as you have already find this.

about inet.*eth0: in this case it is used to replace grep inet | grep eth0, so we need both (inet.*eth0 is order dependent, grep is not). Yes, it is greedy, but it is not really important here.