Why does my Perl one-liner containing an escaped $ not work inside Bash backticks?

Question

I'm trying search for a $ in a text document using a Perl one-liner in a Bash script. I suspect it is treating the $ as a variable.

This is my code:

ITEMVALUE=`perl -ne 'print /\$.*/g' file.txt`

The text looks something like this:

a bunch of random text
$80
a bunch of random text

I want to extract the $80 only. Why isn't this working, and how do I retrieve the desired text?

Check [this](http://stackoverflow.com/questions/9716443/how-to-escape-dollar-sign-in-a-string-using-perl-regex) answer, maybe it helps. — DarkCygnus, Jun 10 '16 at 17:04
The Perl looks fine -- cab we see your entire Bash script, please? — tale852150, Jun 10 '16 at 17:40

score 5 · Accepted Answer · edited Jun 20 '20 at 09:12

From the Bash manual:

Command Substitution

Command substitution allows the output of a command to replace the command name. There are two forms:
$(command)
or
`command` 
...

When the old-style backquote form of substitution is used, backslash retains its literal meaning except when followed by $, `, or \ ... When using the $(command) form, all characters between the parentheses make up the command; none are treated specially.

(emphasis added)

Bash is treating the \ in \$ as an escape sequence, so

ITEMVALUE=`perl -ne 'print /\$.*/g' file.txt`

actually runs

perl -ne 'print /$.*/g' file.txt

$. is a Perl built-in variable representing the current line number, so the regex becomes /1*/ for the first line, /2*/ for the second line, and so on. You'll get output for any line containing its own line number.

You can see this using Perl's re pragma:

$ ITEMVALUE=`perl -Mre=debug -ne 'print /\$.*/g' file.txt`
Compiling REx "1*"
...
Compiling REx "2*"
...

To fix, use the $(command) form of command substitution, which is generally preferred anyway:

ITEMVALUE=$(perl -ne 'print /\$.*/g' file.txt)

Side note: using `/g` is strange since `\$.*` can only match once per line, so I would prefer `print $1 if /(\$.*)/` — ThisSuitIsBlackNot, Jun 10 '16 at 18:49

score 0 · Answer 2 · edited Jun 10 '16 at 17:07

0

It's simpler than that - that doesn't do what you think it does.

You probably want:

perl -ne 'print $1 if m/(\$\d+)/'

Which matches the regex, catches the value, and prints the value. (note - will match multiple times potentially).

edited Jun 10 '16 at 17:07

Borodin

126,100
9
70
144

answered Jun 10 '16 at 16:57

Sobrique

52,974
7
60
101

perhaps allow it to catch a cents value as well if it's present? – stevieb Jun 10 '16 at 16:58
2

Also, that throws a warning: `Possible unintended interpolation of $\ in regex at -e line 1.` – stevieb Jun 10 '16 at 17:07
1

`echo -e 'foo\n$80\nbar' | perl -ne 'print /\$.*/g'` (OP's version) prints `$80`, the same as yours. Why do you say "that doesn't do what you think it does"? – ThisSuitIsBlackNot Jun 10 '16 at 17:08
Your option worked, I found an alternative to my solution. Use ITEMVALUE=$(command) to set variable. – hemlock Jun 10 '16 at 17:32
This solution has one backquote and one regular quote? – mob Jun 10 '16 at 22:44

Why does my Perl one-liner containing an escaped $ not work inside Bash backticks?

2 Answers2

Linked