Sed to replace a speccificc part of line by argument in shell script based in line number

I have a script test.sh
He has some arguments like :

./test.sh Forest Cuba World Hello

JavaScript
​x
 
./test.sh Forest Cuba World Hello​

I made a while read loop to replace each specific part of a line (extractReplaceArgLine) with argument passed in shell script (Forest Cuba World Hello).

while read line
do
    extractReplaceArgLine=$(echo "${line}" | cut -d ' ' -f 1)
    echo "${line}" | sed -e "s/"${extractReplaceArgLine}"/What I put Here ?/"
done < test.txt

JavaScript
 
while read linedo    extractReplaceArgLine=$(echo "${line}" | cut -d ' ' -f 1)    echo "${line}" | sed -e "s/"${extractReplaceArgLine}"/What I put Here ?/"done < test.txt​

Consider the file test.txt :

Ocean is incredible
Spain is my favorite country
Moon calls me everyday
Goodbye World

JavaScript
 
Ocean is incredibleSpain is my favorite countryMoon calls me everydayGoodbye World ​

So, the output would be :

Forest is incredible
Cuba is my favorite country
World calls me everyday
Hello World

JavaScript
 
Forest is incredibleCuba is my favorite countryWorld calls me everydayHello World​

How can I do that ?

Thanks !

Answer

Use $1 to get the first argument, and then use shift to shift the arguments down. Doing this in a loop will process each argument in sequence.

There’s also no need to use cut and sed, you can use the shell’s built-in operations.

while read firstword restofline
do
    echo "$1 $restofline"
    shift
done < test.txt

JavaScript
 
while read firstword restoflinedo    echo "$1 $restofline"    shiftdone < test.txt​

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Answer