Differences between revisions 2 and 8 (spanning 6 versions)
Revision 2 as of 2007-06-26 00:17:35
Size: 1625
Editor: cpe-74-65-28-251
Comment: ${x##*([$' \t\n'])} is wrong -- you don't want to match zero or more of those characters. You want +, for one or more.
Revision 8 as of 2008-07-16 07:31:15
Size: 1107
Editor: pgas
Comment: some precision in case the variable contains newlines, fix italics
Deletions are marked like this. Additions are marked like this.
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There are a few ways to do this -- none of them elegant.

First, the most portable way would be to use sed:
There are a few ways to do this:
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   x=$(echo "$x" | sed -e 's/^ *//' -e 's/ *$//')
   # Note: this only removes spaces. For tabs too:
   x=$(echo "$x" | sed -e $'s/^[ \t]*//' -e $'s/[ \t]*$//')
   # Or possibly, with some systems:
   x=$(echo "$x" | sed -e 's/^[[:space:]]\+//' -e 's/[[:space:]]\+$//')
   #POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var
   EOF
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One can achieve the goal using builtins, although at the moment I'm not sure which shells the following syntax supports: One can also achieve in bash using a herestring
{{{
   # Bash still fails if the variable contains a newline.
   read -r x <<< "$x"
}}}

(note: using IFS=$' \t' read -d "" -r x partially fix the "problem" of the newlines but adds a trailing \n)

There's also a solution using [:glob:extglob]:
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   # Remove leading whitespace:
   while [[ $x = [$' \t\n']* ]]; do x=${x#[$' \t\n']}; done
   # And now trailing:
   while [[ $x = *[$' \t\n'] ]]; do x=${x%[$' \t\n']}; done
}}}

Of course, the preceding example is pretty slow, because it removes one character at a time, in a loop (although it's good enough in practice for most purposes). If you want something a bit fancier, there's a bash-only solution using extglob:

{{{
   # Bash
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   x=${x##+([$' \t\n'])}; x=${x%%+([$' \t\n'])}    x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
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Rather than specify each type of space character yourself, you can use character classes. Two character classes that are useful for matching whitespace are space and blank.

More info: ctype/wctype(3), re_format/regex(7), isspace(3).
This also works in KornShell, without needing the explicit `extglob` setting:
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   shopt -s extglob
   x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}
   shopt -u extglob
   # ksh
   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
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There are many, many other ways to do this. These are not necessarily the most efficient, but they're known to work. There are many, many other ways to do this, using sed for instance:
{{{
   # POSIX, suppress the trailing and leading whitespace on every lines
   x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
}}}

These are not necessarily the best, but they're known to work.

Anchor(faq67)

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

There are a few ways to do this:

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

One can also achieve in bash using a herestring

   # Bash still fails if the variable contains a newline.
   read  -r x <<< "$x"

(note: using IFS=$' \t' read -d "" -r x partially fix the "problem" of the newlines but adds a trailing \n)

There's also a solution using [:glob:extglob]:

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

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

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

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

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

These are not necessarily the best, but they're known to work.

BashFAQ/067 (last edited 2018-11-29 15:32:42 by GreyCat)