我有一个shell脚本与这段代码:
var=`hg st -R "$path"`
if [ -n "$var" ]; then
echo $var
fi
但是条件代码总是执行,因为hg st总是打印至少一个换行符。
是否有一个简单的方法从$var中剥离空白(如PHP中的trim())?
or
有没有处理这个问题的标准方法?
我可以使用sed或AWK,但我认为有一个更优雅的解决方案来解决这个问题。
我有一个shell脚本与这段代码:
var=`hg st -R "$path"`
if [ -n "$var" ]; then
echo $var
fi
但是条件代码总是执行,因为hg st总是打印至少一个换行符。
是否有一个简单的方法从$var中剥离空白(如PHP中的trim())?
or
有没有处理这个问题的标准方法?
我可以使用sed或AWK,但我认为有一个更优雅的解决方案来解决这个问题。
当前回答
# Trim whitespace from both ends of specified parameter
trim () {
read -rd '' $1 <<<"${!1}"
}
# Unit test for trim()
test_trim () {
local foo="$1"
trim foo
test "$foo" = "$2"
}
test_trim hey hey &&
test_trim ' hey' hey &&
test_trim 'ho ' ho &&
test_trim 'hey ho' 'hey ho' &&
test_trim ' hey ho ' 'hey ho' &&
test_trim $'\n\n\t hey\n\t ho \t\n' $'hey\n\t ho' &&
test_trim $'\n' '' &&
test_trim '\n' '\n' &&
echo passed
其他回答
答案有很多,但我仍然认为我刚刚写的剧本值得一提,因为:
it was successfully tested in the shells bash/dash/busybox shell it is extremely small it doesn't depend on external commands and doesn't need to fork (->fast and low resource usage) it works as expected: it strips all spaces and tabs from beginning and end, but not more important: it doesn't remove anything from the middle of the string (many other answers do), even newlines will remain special: the "$*" joins multiple arguments using one space. if you want to trim & output only the first argument, use "$1" instead if doesn't have any problems with matching file name patterns etc
脚本:
trim() {
local s2 s="$*"
until s2="${s#[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
until s2="${s%[[:space:]]}"; [ "$s2" = "$s" ]; do s="$s2"; done
echo "$s"
}
用法:
mystring=" here is
something "
mystring=$(trim "$mystring")
echo ">$mystring<"
输出:
>here is
something<
# Strip leading and trailing white space (new line inclusive).
trim(){
[[ "$1" =~ [^[:space:]](.*[^[:space:]])? ]]
printf "%s" "$BASH_REMATCH"
}
OR
# Strip leading white space (new line inclusive).
ltrim(){
[[ "$1" =~ [^[:space:]].* ]]
printf "%s" "$BASH_REMATCH"
}
# Strip trailing white space (new line inclusive).
rtrim(){
[[ "$1" =~ .*[^[:space:]] ]]
printf "%s" "$BASH_REMATCH"
}
# Strip leading and trailing white space (new line inclusive).
trim(){
printf "%s" "$(rtrim "$(ltrim "$1")")"
}
OR
# Strip leading and trailing specified characters. ex: str=$(trim "$str" $'\n a')
trim(){
if [ "$2" ]; then
trim_chrs="$2"
else
trim_chrs="[:space:]"
fi
[[ "$1" =~ ^["$trim_chrs"]*(.*[^"$trim_chrs"])["$trim_chrs"]*$ ]]
printf "%s" "${BASH_REMATCH[1]}"
}
OR
# Strip leading specified characters. ex: str=$(ltrim "$str" $'\n a')
ltrim(){
if [ "$2" ]; then
trim_chrs="$2"
else
trim_chrs="[:space:]"
fi
[[ "$1" =~ ^["$trim_chrs"]*(.*[^"$trim_chrs"]) ]]
printf "%s" "${BASH_REMATCH[1]}"
}
# Strip trailing specified characters. ex: str=$(rtrim "$str" $'\n a')
rtrim(){
if [ "$2" ]; then
trim_chrs="$2"
else
trim_chrs="[:space:]"
fi
[[ "$1" =~ ^(.*[^"$trim_chrs"])["$trim_chrs"]*$ ]]
printf "%s" "${BASH_REMATCH[1]}"
}
# Strip leading and trailing specified characters. ex: str=$(trim "$str" $'\n a')
trim(){
printf "%s" "$(rtrim "$(ltrim "$1" "$2")" "$2")"
}
OR
建立在moskit的expr soulution…
# Strip leading and trailing white space (new line inclusive).
trim(){
printf "%s" "`expr "$1" : "^[[:space:]]*\(.*[^[:space:]]\)[[:space:]]*$"`"
}
OR
# Strip leading white space (new line inclusive).
ltrim(){
printf "%s" "`expr "$1" : "^[[:space:]]*\(.*[^[:space:]]\)"`"
}
# Strip trailing white space (new line inclusive).
rtrim(){
printf "%s" "`expr "$1" : "^\(.*[^[:space:]]\)[[:space:]]*$"`"
}
# Strip leading and trailing white space (new line inclusive).
trim(){
printf "%s" "$(rtrim "$(ltrim "$1")")"
}
使用Bash的扩展模式匹配功能(shopt -s extglob),你可以这样使用:
修剪{# # * ()}
删除任意数量的前导空格。
如果启用了shop -s extglob,那么下面是一个简洁的解决方案。
这招对我很管用:
text=" trim my edges "
trimmed=$text
trimmed=${trimmed##+( )} #Remove longest matching series of spaces from the front
trimmed=${trimmed%%+( )} #Remove longest matching series of spaces from the back
echo "<$trimmed>" #Adding angle braces just to make it easier to confirm that all spaces are removed
#Result
<trim my edges>
用更少的行数来获得相同的结果:
text=" trim my edges "
trimmed=${${text##+( )}%%+( )}
你可以使用tr删除换行符:
var=`hg st -R "$path" | tr -d '\n'`
if [ -n $var ]; then
echo $var
done