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
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<
其他回答
Python有一个函数strip(),它的工作原理与PHP的trim()相同,因此我们可以使用一点内联Python来创建一个易于理解的实用程序:
alias trim='python -c "import sys; sys.stdout.write(sys.stdin.read().strip())"'
这将删除开头和结尾的空白(包括换行符)。
$ x=`echo -e "\n\t \n" | trim`
$ if [ -z "$x" ]; then echo hi; fi
hi
我总是用sed来做
var=`hg st -R "$path" | sed -e 's/ *$//'`
如果有更优雅的解决方案,我希望有人能发布出来。
剥离一个前导和一个后导空间
trim()
{
local trimmed="$1"
# Strip leading space.
trimmed="${trimmed## }"
# Strip trailing space.
trimmed="${trimmed%% }"
echo "$trimmed"
}
例如:
test1="$(trim " one leading")"
test2="$(trim "one trailing ")"
test3="$(trim " one leading and one trailing ")"
echo "'$test1', '$test2', '$test3'"
输出:
'one leading', 'one trailing', 'one leading and one trailing'
去掉所有前导和尾随空格
trim()
{
local trimmed="$1"
# Strip leading spaces.
while [[ $trimmed == ' '* ]]; do
trimmed="${trimmed## }"
done
# Strip trailing spaces.
while [[ $trimmed == *' ' ]]; do
trimmed="${trimmed%% }"
done
echo "$trimmed"
}
例如:
test4="$(trim " two leading")"
test5="$(trim "two trailing ")"
test6="$(trim " two leading and two trailing ")"
echo "'$test4', '$test5', '$test6'"
输出:
'two leading', 'two trailing', 'two leading and two trailing'
#!/bin/bash
function trim
{
typeset trimVar
eval trimVar="\${$1}"
read trimVar << EOTtrim
$trimVar
EOTtrim
eval $1=\$trimVar
}
# Note that the parameter to the function is the NAME of the variable to trim,
# not the variable contents. However, the contents are trimmed.
# Example of use:
while read aLine
do
trim aline
echo "[${aline}]"
done < info.txt
# File info.txt contents:
# ------------------------------
# ok hello there $
# another line here $
#and yet another $
# only at the front$
#$
# Output:
#[ok hello there]
#[another line here]
#[and yet another]
#[only at the front]
#[]
Trim()删除空白(以及制表符、不可打印字符;为了简单起见,我正在考虑使用空格)。我的解决方案是:
var="$(hg st -R "$path")" # I often like to enclose shell output in double quotes
var="$(echo "${var}" | sed "s/\(^ *\| *\$\)//g")" # This is my suggestion
if [ -n "$var" ]; then
echo "[${var}]"
fi
sed命令只修剪开头和结尾的空白,但它也可以被输送到第一个命令,导致:
var="$(hg st -R "$path" | sed "s/\(^ *\| *\$\)//g")"
if [ -n "$var" ]; then
echo "[${var}]"
fi