Diff for "BashFAQ/067"

Differences between revisions 2 and 7 (spanning 5 versions)

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.

-  ⇤ ← 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 7 as of 2008-07-16 07:29:57 → ⇥
  Size: 1107
  Editor: pgas
  Comment: some precision in case the variable contains newlines
-Deletions are marked like this.
+Additions are marked like this.
 Line 3:
-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:
-Line 8:
+Line 6:
-   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
   $var 
   EOF
-Line 15:
+Line 12:
-One can achieve the goal using builtins, although at the moment I'm not sure which shells the following syntax supports:
+One can also achieve in bash using a herestring
{{{
   # Bash still fails if the variable contains a newline.
   read  -r x <<< "$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]:
-Line 18:
+Line 23:
-   # 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
-Line 28:
+Line 25:
-   x=${x##+([$' \t\n'])}; x=${x%%+([$' \t\n'])}
+   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
-Line 32:
+Line 29:
-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:
-Line 37:
+Line 32:
-   shopt -s extglob
   x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}
   shopt -u extglob
+   # ksh
   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
-Line 42:
+Line 36:
-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.