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| <<Anchor(faq67)>> == 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 [[glob|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 [[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#$x)) # However, this fails if x contains anything other than decimal digits. }}} If you need to remove leading zeroes in a POSIX shell, you can use a loop: {{{ # POSIX while true; do case "$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. |
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My name is Valentine. Alaska is where my house is. Fencing is the only pastime her partner doesn't authorize of. After running out his job for several years he ended up being a hotel receptionist. Have a look at the most recent news on my site: https://www.youtube.com/watch?v=S6joRgSfOxA<<BR>><<BR>>
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