Diff for "BashFAQ/067"

Differences between revisions 3 and 28 (spanning 25 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=${var%%[! ]*}   # remove all but leading spaces
var=${var#"$junk"}   # remove leading spaces from original string

junk=${var##*[! ]}   # remove all but trailing spaces
var=${var%"$junk"}   # remove trailing spaces from original string

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))

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 parameter expansion trick (covered in more detail in FAQ #100):

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

Or this mathematical trick:

# POSIX
a=1$var
b=2$var
var=$((2*a - b))
# This may overflow if $var is very large.

It should be noted that all of the tricks for removing leading zeroes will fail if the input value has a sign (- or +). If you're dealing with potentially signed numeric inputs and need to remove leading zeros to avoid them being interpreted as octal, please see ArithmeticExpression.

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 3 as of 2007-06-27 18:33:16 → 
  Size: 1623
  Editor: GreyCat
  Comment: Remove GNUism (sed s/x\+//).  Ordinary Kleene closure (*) is perfectly adequate here.  (Likewise in the extglob part, but I left that alone.)
+   ← Revision 28 as of 2025-07-19 12:59:21 → ⇥
  Size: 3497
  Editor: GreyCat
  Comment: add a mathematical trick for leading zero removal, normalize whitespace, add caveat about signed inputs
-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 8:
-   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:]]*$//')
+# POSIX
junk=${var%%[! ]*}   # remove all but leading spaces
var=${var#"$junk"}   # remove leading spaces from original string

junk=${var##*[! ]}   # remove all but trailing spaces
var=${var%"$junk"}   # remove trailing spaces from original string
-Line 15:
+Line 16:
-One can achieve the goal using builtins, although at the moment I'm not sure which shells the following syntax supports:
+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 18:
+Line 19:
-   # 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
+# Bash
shopt -s extglob
var=${var##*( )}   # trim the left
var=${var%%*( )}   # trim the right
-Line 24:
+Line 25:
-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:
+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 27:
+Line 28:
-   shopt -s extglob
   x=${x##+([$' \t\n'])}; x=${x%%+([$' \t\n'])}
   shopt -u extglob
+# POSIX, but fails if the variable contains newlines
read -r var << EOF
$var
EOF
-Line 32:
+Line 34:
-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).
+Bash can do something similar with a "here string":
 Line 37:
-   shopt -s extglob
   x=${x##+([[:space:]])}; x=${x%%+([[:space:]])}
   shopt -u extglob
+# Bash
read  -rd '' x <<< "$x"
-Line 42:
+Line 41:
-There are many, many other ways to do this.  These are not necessarily the most efficient, but they're known to work.
+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]]:

{{{
# 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 [[ArithmeticExpression|math context]]:

{{{
# Bash
x=$((10#$x))
}}}

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 parameter expansion trick (covered in more detail in [[BashFAQ/100|FAQ #100]]):

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

Or this mathematical trick:

{{{
# POSIX
a=1$var
b=2$var
var=$((2*a - b))
# This may overflow if $var is very large.
}}}

It should be noted that all of the tricks for removing leading zeroes will '''fail''' if the input value has a sign (`-` or `+`). If you're dealing with potentially signed numeric inputs and need to remove leading zeros to avoid them being interpreted as octal, please see [[ArithmeticExpression]].

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.