Bash append to variable using while read syntax

Question

I have file which looks like following:

/usr/local/bin
/bin
/usr/bin
/usr/local/sbin
/usr/sbin
/sbin
/home/user/.local/bin
/home/user/bin
/home/user/perl5/bin

now I want to concatenate the lines by colon using while read syntax, something like this:

cat file | while read data; do path="$path:$data"; done && echo $path

but above code is not working that I am expecting, the $path variable remains empty, what is the problem?

possible duplicate of Can't read variable that was stored from within a while loop, when out of the while loop — fedorqui
– fedorqui, Commented Sep 23, 2013 at 12:43

fedorqui · Accepted Answer · 2013-09-23 12:38:03Z

2

You can get this output doing the following:

$ tr -s '\n' ':' <file
/usr/local/bin:/bin:/usr/bin:/usr/local/sbin:/usr/sbin:/sbin:/home/user/.local/bin:/home/user/bin:/home/user/perl5/bin

It replaces all new lines with a colon.

Regarding the error in your syntax, note that the proper syntax is:

while read data
do
   ... things ...
done < file

answered Sep 23, 2013 at 12:38

fedorqui

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1 Comment

Armali Over a year ago

+1 for the loop syntax. +0 for the tr command - it indeed replaces all newlines with a colon, including the last, which produces an unwanted trailing colon (that is missing from the above output).

Armali · Accepted Answer · 2013-09-25 06:16:33Z

1

We don't need a loop or an external command.

path=$(<file) && path=${path//$'\n'/:} && echo "$path"

answered Sep 25, 2013 at 6:16

Armali

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