$(tail) into variable removing \n

Question

I am doing the following:

x="Hello to the world of tomorrow\n <pre>";
x="${x}$(tail -n 50 log/logfile.log)"
x="${x}</pre>";
echo -e $x | ./perlscript

The Perl script:

#!perl
# perlscript
open(MAIL, "sendmail -t")
print MAIL "EMAIL HEADERS\n\n"
print MAIL <STDIN>
close(MAIL);

When I receive the email, the log file doesn't have any \n inside the <pre> tag. The Hello to the world of tomorrow works fine and the \n is represented.

How to I get tail not to remove any \n, or is it something further down/up the chain?

Is the first code block a bashscript or something, or are you just typing it on the command line? — Bernhard, Sep 26 '12 at 14:50
Do you actually want `\n` (2 characters) or a newline character? Keep in mind `echo` is very nonportable when it comes to handling backslashes. If you want predictable behavior use `printf` instead of `echo`. — jw013, Sep 26 '12 at 14:50
Yes i Want the `\n`, and its a bashscript, but could be command line — whoami, Sep 26 '12 at 14:51
@whoami No, saying you want `echo -e \n` probably means you *really* want an actual newline, and `echo -e \n` is your attempt at getting one. — jw013, Sep 26 '12 at 15:19
Completely unrelated, but why are you embedding what looks to be HTML tags into a plain text email? That is not good form. — jw013, Sep 26 '12 at 15:32
@jw013 - Its not plain text, `print MAIL "EMAIL HEADERS\n\n"` was an abbreviation for HTML content type header, and all the to: from: subject: fields. Sorry was lazy to type :D — whoami, Sep 26 '12 at 15:50

score 5 · Accepted Answer · answered Sep 26 '12 at 15:01

5

double quotes are missing in the command echo :

echo -e "$x" | ./perlscript

answered Sep 26 '12 at 15:01

Nahuel Fouilleul

1,554
10
8

Thats the one....Would be nice to understand why the differnce between `echo -e "$x"` and `echo -e $x` – whoami Sep 26 '12 at 15:12
2

@whoami, look in the shell man page for "word splitting" -- when you don't quote the variable, all sequences of whitespace are replaced by a single space. – glenn jackman Sep 26 '12 at 15:35
I don't think this is the only problem. I think that `$(...)` constructs also drop the trailing space. Witness that `A=$(printf "abc\n"); printf "<%s>" "$A"` will produce ``. – dubiousjim Sep 26 '12 at 16:03
I agree but this error had to be identified – Nahuel Fouilleul Sep 26 '12 at 16:35
@whoami Always use double quotes around variable substitutions and command substitutions: `"$x"`, `"$(somecommand)"`. See http://mywiki.wooledge.org/BashFAQ/020,[$VAR vs ${VAR} and to quote or not to quote](http://unix.stackexchange.com/q/4899) – Gilles 'SO- stop being evil' Sep 26 '12 at 23:03

score 5 · Answer 2 · answered Sep 26 '12 at 15:22

This is a much cleaner and more idiomatic way to go about what you are trying to do:

{ printf 'Hello to the world of tomorrow\n <pre>\n'
  tail -n 50 log/logfile.log
  printf '</pre>\n'; } | ./perlscript

There is no reason you need to buffer all that output in a variable first when you are just going to push it in order into the pipe anyways.

$(tail) into variable removing \n

2 Answers2