Question

I am trying to write a bash script(script.sh) to search and replace some variables in input.sh file. But I need to modify only the variables which are present in variable_list file and leave others as it is.

variable_list

${user}  
${dbname}

input.sh

username=${user}  
password=${password}  
dbname=${dbname}

Expected output file

username=oracle  
password=${password} > This line won't be changed as this variable(${password}) is not in variable_list file
dbname=oracle

Following is the script I am trying to use but I am not able to find the correct sed expression

script.sh

export user=oracle  
export password=oracle123  
export dbname=oracle

variable='variable_list'  
while read line ;  
do  
 if [[ -n $line ]]     
 then  
  sed -i 's/$line/$line/g' input.sh > output.sh  
 fi  
done < "$variable"    

Answer 1

This could work:

#!/bin/bash

export user=oracle  
export password=oracle123  
export dbname=oracle

variable='variable_list'  
while read line ;  
do  
 if [[ -n $line ]]
 then
  exp=$(sed -e 's/\$/\\&/g' <<< "$line")
  var=$(sed -e 's/\${\([^}]\+\)}/\1/' <<< "$line")
  sed -i "s/$exp/${!var}/g" input.sh
 fi  
done < "$variable"

The first sed expression escapes the $ which is a regex metacharacter. The second extracts just the variable name, then we use indirection to get the value in our current shell and use it in the sed expression.

Edit

Rather than rewriting the file so many times, it’s probably more efficient to do it like this, building the arguments list for sed:

#!/bin/bash

export user=oracle  
export password=oracle123  
export dbname=oracle

while read var
do
    exp=$(sed -e 's/\$/\\&/g' <<< "$var")
    var=$(sed -e 's/\${\([^}]\+\)}/\1/' <<< "$var")
    args+=("-e s/$exp/${!var}/g")
done < "variable_list"

sed "${args[@]}" input.sh > output.sh