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
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。
其他回答
var = ' a b '
# remove all white spaces
new=$(echo $var | tr -d ' ')
# remove leading and trailing whitespaces
new=$(echo $var)
ab
a b
数组赋值在内部字段分隔符(默认为空格/制表符/换行符)上展开参数拆分。
words=($var)
var="${words[@]}"
您可以使用老式的tr。例如，它返回git存储库中已修改文件的数量，删除空白。
MYVAR=`git ls-files -m|wc -l|tr -d ' '`
# 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")")"
}
要从左到第一个单词删除空格和制表符，输入:
echo " This is a test" | sed "s/^[ \t]*//"
cyberciti.biz /技巧/ delete-leading-spaces-from-front-of-each-word.html
