Diff for "BashFAQ/067"

How can I trim leading/trailing white space from one of my variables?

There are a few ways to do this. Some involve special tricks that only work with whitespace. Others are more general, and can be used to strip leading zeroes, etc.

For simple variables, you can trim whitespace (or other characters) using this trick:

   # POSIX
   junk=${var##*[! ]}   # match prefixed spaces
   var=${var%"$junk"}    # trim the match
   junk=${var%%[! ]*}   # match affixed spaces
   var=${var#"$junk"}    # trim the match

Bash can do the same thing, but without the need for a throw-away variable, by using extglob's more advanced pattern matching:

   # Bash
   shopt -s extglob
   var=${var##*( )}   # trim the left
   var=${var%%*( )}   # trim the right

Here's one that only works for whitespace. It relies on the fact that read strips all leading and trailing whitespace (tab or space character) when IFS isn't set:

   # POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var
   EOF

Bash can do something similar with a "here string":

   # Bash
   read  -rd '' x <<< "$x"

Using an empty string as a delimiter means the read consumes the whole string, as NUL is used. (Remember: BASH only does C-string variables.) This is entirely safe for any text, including newlines (which will also be stripped from the beginning and end of the variable with the default value of IFS).

Here's a solution using extglob together with parameter expansion:

   # Bash
   shopt -s extglob
   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}

(where [[:space:]] includes space, tab and all other horizontal and vertical spacing characters, the list of which varies with the current locale).

This also works in KornShell, without needing the explicit extglob setting:

   # ksh
   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}

This solution isn't restricted to whitespace like the first few were. You can remove leading zeroes as well:

   # Bash
   shopt -s extglob
   x=${x##+(0)}

Another way to remove leading zeroes from a number in bash is to treat it as a decimal integer, in a math context:

   # Bash
   x=$((10#$x))
   # However, this fails if x contains anything other than decimal digits.

If you need to remove leading zeroes in a POSIX shell, you can use a loop:

   # POSIX
   while true; do
     case "$var" in
       0*) var=${var#0};;
       *)  break;;
     esac
   done

Or this trick (covered in more detail in FAQ #100):

   # POSIX
   zeroes=${var%%[!0]*}
   var=${var#"$zeroes"}

There are many, many other ways to do this, using sed for instance:

   # POSIX, suppress the trailing and leading whitespace on every line
   x=$(printf '%s\n' "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')

Solutions based on external programs like sed are better suited to trimming large files, rather than shell variables.