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
我将简单地使用sed:
function trim
{
echo "$1" | sed -n '1h;1!H;${;g;s/^[ \t]*//g;s/[ \t]*$//g;p;}'
}
a)单行字符串使用示例
string=' wordA wordB wordC wordD '
trimmed=$( trim "$string" )
echo "GIVEN STRING: |$string|"
echo "TRIMMED STRING: |$trimmed|"
输出:
GIVEN STRING: | wordA wordB wordC wordD |
TRIMMED STRING: |wordA wordB wordC wordD|
b)多行字符串使用示例
string=' wordA
>wordB<
wordC '
trimmed=$( trim "$string" )
echo -e "GIVEN STRING: |$string|\n"
echo "TRIMMED STRING: |$trimmed|"
输出:
GIVEN STRING: | wordAA
>wordB<
wordC |
TRIMMED STRING: |wordAA
>wordB<
wordC|
c)最后说明: 如果你不喜欢使用函数,对于单行字符串,你可以简单地使用“更容易记住”的命令,比如:
echo "$string" | sed -e 's/^[ \t]*//' | sed -e 's/[ \t]*$//'
例子:
echo " wordA wordB wordC " | sed -e 's/^[ \t]*//' | sed -e 's/[ \t]*$//'
输出:
wordA wordB wordC
在多行字符串上使用上述方法也可以,但请注意,它也会切断任何尾随/前导内部多行空间,就像GuruM在评论中注意到的那样
string=' wordAA
>four spaces before<
>one space before< '
echo "$string" | sed -e 's/^[ \t]*//' | sed -e 's/[ \t]*$//'
输出:
wordAA
>four spaces before<
>one space before<
所以如果你介意保留这些空格,请使用我回答开头的函数!
d)解释sed语法“find and replace”在函数trim中使用的多行字符串:
sed -n '
# If the first line, copy the pattern to the hold buffer
1h
# If not the first line, then append the pattern to the hold buffer
1!H
# If the last line then ...
$ {
# Copy from the hold to the pattern buffer
g
# Do the search and replace
s/^[ \t]*//g
s/[ \t]*$//g
# print
p
}'
其他回答
为了删除字符串开头和结尾的所有空格(包括行末字符):
echo $variable | xargs echo -n
这也将删除重复的空格:
echo " this string has a lot of spaces " | xargs echo -n
Produces: 'this string has a lot of spaces'
你可以使用tr删除换行符:
var=`hg st -R "$path" | tr -d '\n'`
if [ -n $var ]; then
echo $var
done
# 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")")"
}
我创建了以下函数。我不确定printf的可移植性如何,但这个解决方案的美妙之处在于,您可以通过添加更多字符代码来指定什么是“空白”。
iswhitespace()
{
n=`printf "%d\n" "'$1'"`
if (( $n != "13" )) && (( $n != "10" )) && (( $n != "32" )) && (( $n != "92" )) && (( $n != "110" )) && (( $n != "114" )); then
return 0
fi
return 1
}
trim()
{
i=0
str="$1"
while (( i < ${#1} ))
do
char=${1:$i:1}
iswhitespace "$char"
if [ "$?" -eq "0" ]; then
str="${str:$i}"
i=${#1}
fi
(( i += 1 ))
done
i=${#str}
while (( i > "0" ))
do
(( i -= 1 ))
char=${str:$i:1}
iswhitespace "$char"
if [ "$?" -eq "0" ]; then
(( i += 1 ))
str="${str:0:$i}"
i=0
fi
done
echo "$str"
}
#Call it like so
mystring=`trim "$mystring"`
Bash有一个称为参数展开的特性,它允许基于所谓的模式替换字符串(模式类似于正则表达式,但有基本的区别和限制)。 [flussence的原文:Bash有正则表达式,但它们隐藏得很好:]
下面演示如何从变量值中删除所有空白(甚至来自内部)。
$ var='abc def'
$ echo "$var"
abc def
# Note: flussence's original expression was "${var/ /}", which only replaced the *first* space char., wherever it appeared.
$ echo -n "${var//[[:space:]]/}"
abcdef