Diff for "BashFAQ/067"

Differences between revisions 2 and 6 (spanning 4 versions)

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

There are a few ways to do this

   #POSIX
   read -r var << EOF
   $var 
   EOF

One can also achieve in bash (and ksh93) using a herestring

   # Bash
   # Remove leading whitespace:
   read -r x <<< "$x"

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
   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 6 as of 2008-07-04 09:01:07 → ⇥
  Size: 883
  Editor: pgas
  Comment: posix solution with a here string
-Deletions are marked like this.
+Additions are marked like this.
 Line 3:
-There are a few ways to do this -- none of them elegant.
+There are a few ways to do this 
{{{
   #POSIX
   read -r var << EOF
   $var 
   EOF 
}}}
-Line 5:
+Line 11:
-First, the most portable way would be to use sed:
+One can also achieve in bash (and ksh93) using a herestring
-Line 8:
+Line 14:
-   x=$(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:]]\+$//')
+   # Bash
   # Remove leading whitespace:
   read -r x <<< "$x"
-Line 15:
+Line 19:
-One can achieve the goal using builtins, although at the moment I'm not sure which shells the following syntax supports:
+There's also a solution using [:glob:extglob]:
-Line 18:
+Line 22:
-   # 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 24:
-   x=${x##+([$' \t\n'])}; x=${x%%+([$' \t\n'])}
+   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
-Line 32:
+Line 28:
-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 31:
-   shopt -s extglob
   x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}
   shopt -u extglob
+   # ksh
   x=${x##+([[:space:]])} x=${x%%+([[:space:]])}
-Line 42:
+Line 35:
-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
   x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
}}}
These are not necessarily the best, but they're known to work.