Assign the output to variable in shell script

Question

I am new to shell scripting and i am trying to assign the output of this line to a variable but all efforts are in vain.

var2=$(cat "filename")
var1=$(echo " $var2 +3 " | bc )

var2 is properly read from the file and also the output shows the value of the sumation, but the value is not assigned to the var1

ps : Filename contains a single entry which is a number

please edit you Q to include what is in the file `filename` and expected output. More importantly, your shell has modern cmd-substitution (i.e. `$(....)`), then it also has arithmetic substitution, i.e. `var2=$(( $a + $b / $c))` and just plain arithmetic evaluations with `if (( $a > 5)) ; then ...` so generally, you don't need `bc`. (yes there are cases where you might). (and in those arithmetic tests, you don't even need the leading `$` chars (unless like $1)). I have not downvoted. Good luck! — shellter, Dec 18 '15 at 12:23
When you run that snippet of code directly (and just that code) you see the output from `bc` displayed to the terminal directly and `var1` is empty? Because that doesn't sound right. — Etan Reisner, Dec 18 '15 at 12:28
`var1=$(echo " $var2 +3 " | bc ) ; echo $var1 ` has an output which looks so `4.223 ` there is only a empty space when i try to output the value @EtanReisner — vigneshhegde, Dec 18 '15 at 12:38
What about if you add `set -x` to the top and add `declare -p var2 var1` at the end? What does the complete output from that look like? — Etan Reisner, Dec 18 '15 at 12:49

score 1 · Answer 1 · answered Dec 18 '15 at 12:46

1

The code worked when i tried this

var3="$("echo " $var2 +3 " |bc)"

answered Dec 18 '15 at 12:46

vigneshhegde

Seriously? That should break, because you are trying to run a command named `echo `, not `echo`. – chepner Dec 18 '15 at 21:18

score 0 · Answer 2 · edited May 23 '17 at 11:52

0

You are missing parenthesis

var2=$(cat "file")
var3=$(( var2 + 3 ))
echo $var3

Also you can try with expr Something like this

var2=$(cat "filename")
var3=`expr $var2 + 3`
echo $var3

Here is a good answer HIH

edited May 23 '17 at 11:52

Community

answered Dec 18 '15 at 12:21

antorqs

The code the OP has should be working fine. Replacing the sub-shell, pipe and invocation of `bc` by using the shells arithmetic expansion (i.e. `$((...))`) is a good idea but not a material distinction in terms of "fixing" the problem. – Etan Reisner Dec 18 '15 at 12:29
It's actually a bad idea if `var2`, as it seems from the OP's comment, has a floating-point value. – chepner Dec 18 '15 at 21:18

2 Answers2