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Command substitution
$( <COMMANDS> )
` <COMMANDS> `
The command substitution expands to the output of commands. These
commands are executed in a subshell, and their stdout
data is what the
substitution syntax expands to.
All trailing newlines are removed (below is an example for a workaround).
In later steps, if not quoted, the results undergo word
splitting and pathname
expansion. You have to remember that, because
the word splitting will also remove embedded newlines and other IFS
characters and break the results up into several words. Also you'll
probably get unexpected pathname matches. If you need the literal
results, quote the command substitution!
The second form `COMMAND`
is more or less obsolete for Bash, since
it has some trouble with nesting ("inner" backticks need to be
escaped) and escaping characters. Use $(COMMAND)
, it's also POSIX!
When you call an explicit subshell
(COMMAND)
inside the command substitution $()
, then take care, this
way is wrong:
$((COMMAND))
Why? because it collides with the syntax for arithmetic
expansion. You need to separate the command
substitution from the inner (COMMAND)
:
$( (COMMAND) )
Specialities
When the inner command is only an input redirection, and nothing else, for example
$( <FILE )
# or
` <FILE `
then Bash attempts to read the given file and act just if the given
command was cat FILE
.
A closer look at the two forms
In general you really should only use the form $()
, it's
escaping-neutral, it's nestable, it's also POSIX. But take a look at
the following code snips to decide yourself which form you need under
specific circumstances:
[Nesting]{.underline}
Backtick form `...`
is not directly nestable. You will have to
escape the "inner" backticks. Also, the deeper you go, the more escape
characters you need. Ugly.
echo `echo `ls`` # INCORRECT
echo `echo \`ls\`` # CORRECT
echo $(echo $(ls)) # CORRECT
[Parsing]{.underline}
All is based on the fact that the backquote-form is simple character
substitution, while every $()
-construct opens an own, subsequent
parsing step. Everything inside $()
is interpreted as if written
normal on a commandline. No special escaping of nothing is needed:
echo "$(echo "$(ls)")" # nested double-quotes - no problem
[Constructs you should avoid]{.underline}
It's not all shiny with $()
, at least for my current Bash
(3.1.17(1)-release
. :!: [Update: Fixed since 3.2-beta
together
with a misinterpretion of '))' being recognized as arithmetic
expansion [by redduck666]]{.underline}). This command seems to
incorrectly close the substitution step and echo prints "ls" and
")":
echo $(
# some comment ending with a )
ls
)
It seems that every closing ")" confuses this construct. Also a (very uncommon ;-)) construct like:
echo $(read VAR; case "$var" in foo) blah ;; esac) # spits out some error, when it sees the ";;"
# fixes it:
echo $(read VAR; case "$var" in (foo) blah ;; esac) # will work, but just let it be, please ;-)
[Conclusion:]{.underline}
In general, the $()
should be the preferred method:
- it's clean syntax
- it's intuitive syntax
- it's more readable
- it's nestable
- its inner parsing is separate
Examples
To get the date:
DATE="$(date)"
To copy a file and get cp
error output:
COPY_OUTPUT="$(cp file.txt /some/where 2>&1)"
Attention: Here, you need to redirect cp
STDERR
to its STDOUT
target, because command substitution only catches STDOUT
!
Catch stdout and preserve trailing newlines:
var=$(echo -n $'\n'); echo -n "$var"; # $var == ""
var=$(echo -n $'\n'; echo -n x); var="${var%x}"; echo -n "$var" # $var == "\n"
This adds "x" to the output, which prevents the trailing newlines of the previous commands' output from being deleted by $().
By removing this "x" later on, we are left with the previous commands' output with its trailing newlines.
See also
- Internal: Introduction to expansion and substitution
- Internal: Obsolete and deprecated syntax