Bash - how to put each line within quotation

Question

I want to put each line within quotation marks, such as:

abcdefg hijklmn opqrst 

convert to:

"abcdefg" "hijklmn" "opqrst" 

How to do this in Bash shell script?

Answer 1

Using awk

awk '{ print "\""$0"\""}' inputfile 

Using pure bash

while read FOO; do    echo -e "\"$FOO\"" done < inputfile 

where inputfile would be a file containing the lines without quotes.

If your file has empty lines, awk is definitely the way to go:

awk 'NF { print "\""$0"\""}' inputfile 

NF tells awk to only execute the print command when the Number of Fields is more than zero (line is not empty).

Answer 2

I use the following command:

xargs -I{lin} echo \"{lin}\" < your_filename 

The xargs take standard input (redirected from your file) and pass one line a time to {lin} placeholder, and then execute the command at next, in this case a echo with escaped double quotes.

You can use the -i option of xargs to omit the name of the placeholder, like this:

xargs -i echo \"{}\" < your_filename 

In both cases, your IFS must be at default value or with '\n' at least.