Why does echo `cat file` lose all formatting?

Question

When I run cat directly I get nice formatted text exactly as in the file.

$ cat hostnamectl.output
   Static hostname: linuxtb3
         Icon name: computer-vm
           Chassis: vm
        Machine ID: 76073c1aa8cf48ed900c39d1992fbb73
           Boot ID: 24db61d58fd5491bbf82a4bb743e5b72
    Virtualization: kvm
  Operating System: CentOS Linux 7 (Core)
       CPE OS Name: cpe:/o:centos:centos:7
            Kernel: Linux 3.10.0-957.21.3.el7.x86_64
      Architecture: x86-64

When I run it after echo all formatting is lost.

$ echo `cat hostnamectl.output`
Static hostname: linuxtb3 Icon name: computer-vm Chassis: vm Machine ID: 76073c1aa8cf48ed900c39d1992fbb73 Boot ID: 24db61d58fd5491bbf82a4bb743e5b72 Virtualization: kvm Operating System: CentOS Linux 7 (Core) CPE OS Name: cpe:/o:centos:centos:7 Kernel: Linux 3.10.0-957.21.3.el7.x86_64 Architecture: x86-64

What is the reason for such behavior?

It has nothing to do with script vs. command-line so I removed that aspect from your question. — John Kugelman, Jul 25 '19 at 13:00
You will find the answer @[Capturing multiple line output into a Bash variable](https://stackoverflow.com/a/613580/6771046). — U880D, Jul 25 '19 at 13:05
Put doublequotes around it. The formatting was there, you threw it away. Also, prefer `$( )` to backticks, for several reasons.... — Paul Hodges, Jul 25 '19 at 13:21

score 2 · Answer 1 · answered Jul 25 '19 at 13:09

From Greg's Bash Guide:

2. Quoting

Word Splitting is the demon inside Bash that is out to get unsuspecting newcomers or even veterans who let down their guard. ...The best way to protect yourself from this beast is to quote all your strings. Quotes keep your strings in one piece and prevent Word Splitting from tearing them open.

...

Here's another example to show you just how badly you can break things if you neglect to quote when necessary:
$ echo "$(ls -al)"
total 8
drwxr-xr-x   4 lhunath users 1 2007-06-28 13:13 "."/
drwxr-xr-x 102 lhunath users 9 2007-06-28 13:13 ".."/
-rw-r--r--   1 lhunath users 0 2007-06-28 13:13 "a"
-rw-r--r--   1 lhunath users 0 2007-06-28 13:13 "b"
-rw-r--r--   1 lhunath users 0 2007-06-28 13:13 "c"
drwxr-xr-x   2 lhunath users 1 2007-06-28 13:13 "d"/
drwxr-xr-x   2 lhunath users 1 2007-06-28 13:13 "e"/
$ echo $(ls -al)
total 8 drwxr-xr-x 4 lhunath users 1 2007-06-28 13:13 "."/ drwxr-xr-x 102 lhunath users 9 2007-06-28 13:13 ".."/ -rw-r--r-- 1 lhunath users 0 2007-06-28 13:13 "a" -rw-r--r-- 1 lhunath users 0 2007-06-28 13:13 "b" -rw-r--r-- 1 lhunath users 0 2007-06-28 13:13 "c" drwxr-xr-x 2 lhunath users 1 2007-06-28 13:13 "d"/ drwxr-xr-x 2 lhunath users 1 2007-06-28 13:13 "e"/

Just like in the ls -al example, double quotes will fix your problem:

$ echo "`cat hostnamectl.output`"

It is, however, an anti-pattern to capture a command's output with backticks and then immediately echo the captured output. You should simply call the command directly and skip the extra steps:

$ cat hostnamectl.output

Zang MingJie · Answer 2 · 2019-07-25T13:05:12.533

1

When you run:

echo `cat xxx`

You actually calling:

echo line1word1 line1word2 \n line2word1 line1word2 \n line3word1 line3word2

Which is essentially the same as:

echo line1word1 line1word2 line2word1 line1word2 line3word1 line3word2

This how bash handle args, so you get:

line1word1 line1word2 line2word1 line1word2 line3word1 line3word2

BTW, you can use:

echo "`cat xxx`"

where the cat xxx part will be passed to echo as a single string arg1, so you will get expected output

edited Jul 25 '19 at 13:05

answered Jul 25 '19 at 13:02

Zang MingJie

5,164
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if you get `echo line1word1 line1word2 \n line2word1 line1word2 \n line3word1 line3word2` then surely you'd be able to get it interpreted with `echo -e`? – itChi Jul 25 '19 at 13:04
@itChi This is only helping you understand what's happening, not what actually happened, understand it wisely. – Zang MingJie Jul 25 '19 at 13:06

Why does echo `cat file` lose all formatting?

2 Answers2

2. Quoting