How to trim whitespace from a Bash variable?
How to trim whitespace from a Bash variable?
Question
I have a shell script with this code:
var=`hg st -R "$path"`
if [ -n "$var" ]; then
echo $var
fi
But the conditional code always executes, because hg st
always prints at least one newline character.
- Is there a simple way to strip whitespace from
$var
(liketrim()
in PHP)?
or
- Is there a standard way of dealing with this issue?
I could use sed or AWK, but I'd like to think there is a more elegant solution to this problem.
Popular Answer
Let's define a variable containing leading, trailing, and intermediate whitespace:
FOO=' test test test '
echo -e "FOO='${FOO}'"
# > FOO=' test test test '
echo -e "length(FOO)==${#FOO}"
# > length(FOO)==16
How to remove all whitespace (denoted by [:space:]
in tr
):
FOO=' test test test '
FOO_NO_WHITESPACE="$(echo -e "${FOO}" | tr -d '[:space:]')"
echo -e "FOO_NO_WHITESPACE='${FOO_NO_WHITESPACE}'"
# > FOO_NO_WHITESPACE='testtesttest'
echo -e "length(FOO_NO_WHITESPACE)==${#FOO_NO_WHITESPACE}"
# > length(FOO_NO_WHITESPACE)==12
How to remove leading whitespace only:
FOO=' test test test '
FOO_NO_LEAD_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//')"
echo -e "FOO_NO_LEAD_SPACE='${FOO_NO_LEAD_SPACE}'"
# > FOO_NO_LEAD_SPACE='test test test '
echo -e "length(FOO_NO_LEAD_SPACE)==${#FOO_NO_LEAD_SPACE}"
# > length(FOO_NO_LEAD_SPACE)==15
How to remove trailing whitespace only:
FOO=' test test test '
FOO_NO_TRAIL_SPACE="$(echo -e "${FOO}" | sed -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_TRAIL_SPACE='${FOO_NO_TRAIL_SPACE}'"
# > FOO_NO_TRAIL_SPACE=' test test test'
echo -e "length(FOO_NO_TRAIL_SPACE)==${#FOO_NO_TRAIL_SPACE}"
# > length(FOO_NO_TRAIL_SPACE)==15
How to remove both leading and trailing spaces--chain the sed
s:
FOO=' test test test '
FOO_NO_EXTERNAL_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_EXTERNAL_SPACE='${FOO_NO_EXTERNAL_SPACE}'"
# > FOO_NO_EXTERNAL_SPACE='test test test'
echo -e "length(FOO_NO_EXTERNAL_SPACE)==${#FOO_NO_EXTERNAL_SPACE}"
# > length(FOO_NO_EXTERNAL_SPACE)==14
Alternatively, if your bash supports it, you can replace echo -e "${FOO}" | sed ...
with sed ... <<<${FOO}
, like so (for trailing whitespace):
FOO_NO_TRAIL_SPACE="$(sed -e 's/[[:space:]]*$//' <<<${FOO})"
Read more… Read less…
A simple answer is:
echo " lol " | xargs
Xargs will do the trimming for you. It's one command/program, no parameters, returns the trimmed string, easy as that!
Note: this doesn't remove all internal spaces so "foo bar"
stays the same; it does NOT become "foobar"
. However, multiple spaces will be condensed to single spaces, so "foo bar"
will become "foo bar"
. In addition it doesn't remove end of lines characters.
There is a solution which only uses Bash built-ins called wildcards:
var=" abc "
# remove leading whitespace characters
var="${var#"${var%%[![:space:]]*}"}"
# remove trailing whitespace characters
var="${var%"${var##*[![:space:]]}"}"
printf '%s' "===$var==="
Here's the same wrapped in a function:
trim() {
local var="$*"
# remove leading whitespace characters
var="${var#"${var%%[![:space:]]*}"}"
# remove trailing whitespace characters
var="${var%"${var##*[![:space:]]}"}"
printf '%s' "$var"
}
You pass the string to be trimmed in quoted form. e.g.:
trim " abc "
A nice thing about this solution is that it will work with any POSIX-compliant shell.
Reference
Bash has a feature called parameter expansion, which, among other things, allows string replacement based on so-called patterns (patterns resemble regular expressions, but there are fundamental differences and limitations). [flussence's original line: Bash has regular expressions, but they're well-hidden:]
The following demonstrates how to remove all white space (even from the interior) from a variable value.
$ var='abc def'
$ echo "$var"
abc def
# Note: flussence's original expression was "${var/ /}", which only replaced the *first* space char., wherever it appeared.
$ echo -n "${var//[[:space:]]/}"
abcdef
In order to remove all the spaces from the beginning and the end of a string (including end of line characters):
echo $variable | xargs echo -n
This will remove duplicate spaces also:
echo " this string has a lot of spaces " | xargs echo -n
Produces: 'this string has a lot of spaces'
Strip one leading and one trailing space
trim()
{
local trimmed="$1"
# Strip leading space.
trimmed="${trimmed## }"
# Strip trailing space.
trimmed="${trimmed%% }"
echo "$trimmed"
}
For example:
test1="$(trim " one leading")"
test2="$(trim "one trailing ")"
test3="$(trim " one leading and one trailing ")"
echo "'$test1', '$test2', '$test3'"
Output:
'one leading', 'one trailing', 'one leading and one trailing'
Strip all leading and trailing spaces
trim()
{
local trimmed="$1"
# Strip leading spaces.
while [[ $trimmed == ' '* ]]; do
trimmed="${trimmed## }"
done
# Strip trailing spaces.
while [[ $trimmed == *' ' ]]; do
trimmed="${trimmed%% }"
done
echo "$trimmed"
}
For example:
test4="$(trim " two leading")"
test5="$(trim "two trailing ")"
test6="$(trim " two leading and two trailing ")"
echo "'$test4', '$test5', '$test6'"
Output:
'two leading', 'two trailing', 'two leading and two trailing'