How to trim whitespace from a Bash variable?

 

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 (like trim() 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.

    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 "${FOO}" | sed ... echo -e "${FOO}" | sed ... with sed ... <<<${FOO} , like so (for trailing whitespace): 

    FOO_NO_TRAIL_SPACE="$(sed -e 's/[[:space:]]*$//' <<<${FOO})"

    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 the internal spaces so "foo bar" stays the same. It does NOT become "foobar" .

    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:]]}"}"   
echo "===$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%"${var##*[![:space:]]}"}"   
    echo -n "$var"
}

    You pass the string to be trimmed in quoted form. eg: 

    trim "   abc   "

    A nice thing about this solution is that it will work with any POSIX-compliant shell.

    Reference

  • Remove leading & trailing whitespace from a Bash variable (original source)
    • 链接地址: http://www.djcxy.com/p/57076.html

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