Diff for "BashFAQ/067"

Differences between revisions 4 and 23 (spanning 19 versions)

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.

-  ⇤ ← Revision 4 as of 2008-05-21 20:52:00 → 
  Size: 1195
  Editor: GreyCat
  Comment: clean up
+   ← Revision 23 as of 2016-03-23 09:21:21 → ⇥
  Size: 3145
  Editor: geirha
  Comment: empty line in the wrong place
-Deletions are marked like this.
+Additions are marked like this.
 Line 1:
-[[Anchor(faq67)]]
+<<Anchor(faq67)>>
 Line 3:
-There are a few ways to do this -- none of them elegant.
+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.
 Line 5:
-First, the most portable way would be to use `sed`:
+For simple variables, you can trim whitespace (or other characters) using this trick:
 Line 9:
-   x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
+   junk="${var##*[! ]}"   # match prefixed spaces
   var="${var%"$junk"}"    # trim the match
   junk="${var%%[! ]*}"   # match affixed spaces
   var="${var#"$junk"}"    # trim the match
-Line 11:
+Line 14:
-One can also achieve the same goal using Bash builtins:
+Bash can do the same thing, but without the need for a throw-away variable, by using [[glob|extglob]]'s more advanced pattern matching:
-Line 16:
+Line 18:
-   # Remove leading whitespace:
   while [[ $x = [[:space:]]* ]]; do x=${x#[[:space:]]}; done
+   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:
-Line 19:
+Line 24:
-   # And now trailing:
   while [[ $x = *[[:space:]] ]]; do x=${x%[[:space:]]}; done
+{{{
   # POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var
   EOF
-Line 22:
+Line 30:
+Bash can do something similar with a "here string":
-Line 23:
+Line 32:
-Of course, the preceding example is not optimal, because it removes one character at a time, in a loop (although it's good enough in practice for most purposes).  If you want something a bit fancier, there's a solution using [:glob:extglob]:
+{{{
   # Bash
   read  -rd '' x <<< "$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 [[glob|extglob]] together with  [[BashFAQ/073|parameter expansion]]:
-Line 29:
+Line 44:
-   shopt -u extglob
-Line 31:
+Line 45:
+(where `[[:space:]]` includes space, tab and all other horizontal and vertical spacing characters, the list of which varies with the current [[locale]]).
-Line 38:
+Line 53:
+This solution isn't restricted to whitespace like the first few were.  You can remove leading zeroes as well:
-Line 39:
+Line 55:
-There are many, many other ways to do this.  These are not necessarily the best, but they're known to work.
+{{{
   # 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 [[ArithmeticExpression|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 [[BashFAQ/100|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.