List = 1 2 3 4 5
for i in $LIST
do
sum=`expr $l_$i + $b`
done
here l_1, l_2, l_3, l_4 and l_5 are variables having some values.
when I am trying to access those values within the for loop using loop variable i, the value of i is getting substituted in the expression i.e the statement (1 + $b) is getting executed instead of ($l_1 + $b). I have tried various combinations of $ symbol and also used paranthesis but nothing seems to be working.
Please help!!
Instead of using l_i or list as strings, use arrays:
#! /bin/bash
b=11
l=(12 24 46 68 92)
for i in "${l[@]}" ; do
((sum=i+b))
echo $sum
done
You can use two methods if you want to maintain your given example:
eval expr \${L_$i} + $bREF=L_$i; expr ${!REF} + $b;As choroba mentioned bash is case sensitive so please be careful with your variable names.
x=v v=20 i=$((10+$x)) is run, $i will equal 30. There's an answer here which demonstrates this given the example code. And your eval, by the way, is dangerous. Not because eval is evil or any of that nonsense - it's because you're not protecting the order of expansion.
l_1=10 l_2=20 l_3=30 l_4=40 l_5=50
b=100
for i in 1 2 3 4 5
do echo "$((sum=l_$i+b))"
done
The value of $sum is set above in the current shell - it can be expanded or modified in the normal way. This occurs without the subshell/pipe overhead necessary for invoking/forking expr and its parent shell for each iteration.
OUTPUT:
110
120
130
140
150
Or, if you really want to use expr:
for i in 1 2 3 4 5
do sum=`expr "$((l_$i))" + "$b"`
echo "$sum"
done
OUTPUT:
110
120
130
140
150
I'm no bash expert but here's what I came up with
#!/bin/bash
b=1
l_1=1
l_2=2
l_3=3
for i in '$l_'{1,2,3}
do
eval dvar="${i}"
sum=$(($dvar + $b))
done
The i variable of the for loop will be initialized as '$l_1', '$l_2', and so on. Calling eval will then evaluate the value of i into dvar which then can be used in an arithmetic expression.
List = 1 2 3 4 5calls the commandListwith 6 arguments,=being the first of them.$Listand$LISTare two different variables.