这就是我所做的，结果完美而简单:

the_string="        test"
the_string=`echo $the_string`
echo "$the_string"

输出:

test

您可以使用老式的tr。例如，它返回git存储库中已修改文件的数量，删除空白。

MYVAR=`git ls-files -m|wc -l|tr -d ' '`

#!/bin/bash

function trim
{
    typeset trimVar
    eval trimVar="\${$1}"
    read trimVar << EOTtrim
    $trimVar
EOTtrim
    eval $1=\$trimVar
}

# Note that the parameter to the function is the NAME of the variable to trim, 
# not the variable contents.  However, the contents are trimmed.


# Example of use:
while read aLine
do
    trim aline
    echo "[${aline}]"
done < info.txt



# File info.txt contents:
# ------------------------------
# ok  hello there    $
#    another  line   here     $
#and yet another   $
#  only at the front$
#$



# Output:
#[ok  hello there]
#[another  line   here]
#[and yet another]
#[only at the front]
#[]

#Execute this script with the string argument passed in double quotes !! 
#var2 gives the string without spaces.
#$1 is the string passed in double quotes
#!/bin/bash
var2=`echo $1 | sed 's/ \+//g'`
echo $var2

当IFS变量被设置为其他变量时，我需要从脚本中删除空白。依赖Perl实现了这个目的:

# trim() { echo $1; } # This doesn't seem to work, as it's affected by IFS

trim() { echo "$1" | perl -p -e 's/^\s+|\s+$//g'; }

strings="after --> , <-- before,  <-- both -->  "

OLD_IFS=$IFS
IFS=","
for str in ${strings}; do
  str=$(trim "${str}")
  echo "str= '${str}'"
done
IFS=$OLD_IFS

var='   a b c   '
trimmed=$(echo $var)