I have this Bash script and I had a problem in line 16. How can I take the previous result of line 15 and add it to the variable in line 16?
#!/bin/bash
num=0
metab=0
for ((i=1; i<=2; i++)); do
for j in `ls output-$i-*`; do
echo "$j"
metab=$(cat $j|grep EndBuffer|awk '{sum+=$2} END { print sum/120}') (line15)
num= $num + $metab (line16)
done
echo "$num"
done
For integers:
Use arithmetic expansion: $((EXPR))
num=$((num1 + num2))
num=$(($num1 + $num2)) # Also works
num=$((num1 + 2 + 3)) # ...
num=$[num1+num2] # Old, deprecated arithmetic expression syntax
Using the external expr utility. Note that this is only needed for really old systems.
num=`expr $num1 + $num2` # Whitespace for expr is important
For floating point:
Bash doesn't directly support this, but there are a couple of external tools you can use:
num=$(awk "BEGIN {print $num1+$num2; exit}")
num=$(python -c "print $num1+$num2")
num=$(perl -e "print $num1+$num2")
num=$(echo $num1 + $num2 | bc) # Whitespace for echo is important
You can also use scientific notation (for example, 2.5e+2).
Common pitfalls:
When setting a variable, you cannot have whitespace on either side of =, otherwise it will force the shell to interpret the first word as the name of the application to run (for example, num= or num)
num= 1num =2
bc and expr expect each number and operator as a separate argument, so whitespace is important. They cannot process arguments like 3+ +4.
num=`expr $num1+ $num2`