Diff for "BashFAQ/067"

Differences between revisions 10 and 27 (spanning 17 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))
   # 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 10 as of 2008-07-17 15:56:25 → 
  Size: 1107
  Editor: wsip-68-15-32-50
  Comment:
+   ← Revision 27 as of 2018-11-29 15:32:42 → ⇥
  Size: 3195
  Editor: GreyCat
  Comment: Fix comments on first example, because they were backwards and misleading.  Also reverse the steps because humans expect leading before trailing.
-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:
+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:
-Line 6:
+Line 8:
-   #POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var 
   EOF
+# POSIX
junk=${var%%[! ]*}   # remove all but leading spaces
var=${var#"$junk"}   # remove leading spaces from original string

junk=${var##*[! ]}   # remove all but trailing spaces
var=${var%"$junk"}   # remove trailing spaces from original string
-Line 12:
+Line 16:
-One can also achieve in bash using a herestring
+Bash can do the same thing, but without the need for a throw-away variable, by using [[glob|extglob]]'s more advanced pattern matching:
-Line 14:
+Line 19:
-   # Bash still fails if the variable contains a newline.
   read  -r x <<< "$x"
+   # Bash
   shopt -s extglob
   var=${var##*( )}   # trim the left
   var=${var%%*( )}   # trim the right
-Line 17:
+Line 24:
+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 18:
+Line 26:
-(note: using IFS=$' \t' read -d "" -r x partially fix the "problem" of the newlines but adds a trailing \n)
+{{{
   # POSIX, but fails if the variable contains newlines
   read -r var << EOF
   $var
   EOF
}}}
Bash can do something similar with a "here string":
-Line 20:
+Line 34:
-There's also a solution using [:glob:extglob]:
+{{{
   # Bash
   read  -rd '' x <<< "$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 26:
+Line 46:
-   shopt -u extglob
-Line 28:
+Line 47:
+(where `[[:space:]]` includes space, tab and all other horizontal and vertical spacing characters, the list of which varies with the current [[locale]]).
-Line 35:
+Line 55:
+This solution isn't restricted to whitespace like the first few were.  You can remove leading zeroes as well:
-Line 36:
+Line 57:
-There are many, many other ways to do this, using sed for instance:
-Line 38:
+Line 58:
-   # POSIX, suppress the trailing and leading whitespace on every lines
   x=$(echo "$x" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')
+   # Bash
   shopt -s extglob
   x=${x##+(0)}
-Line 41:
+Line 62:
+Another way to remove leading zeroes from a number in bash is to treat it as a decimal integer, in a [[ArithmeticExpression|math context]]:
-Line 42:
+Line 64:
-These are not necessarily the best, but they're known to work.
+{{{
   # 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.