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
让我们定义一个包含开头、结尾和中间空格的变量:
FOO=' test test test '
echo -e "FOO='${FOO}'"
# > FOO=' test test test '
echo -e "length(FOO)==${#FOO}"
# > length(FOO)==16
如何删除tr中的所有空格(由[:space:]表示):
FOO=' test test test '
FOO_NO_WHITESPACE="$(echo -e "${FOO}" | tr -d '[:space:]')"
echo -e "FOO_NO_WHITESPACE='${FOO_NO_WHITESPACE}'"
# > FOO_NO_WHITESPACE='testtesttest'
echo -e "length(FOO_NO_WHITESPACE)==${#FOO_NO_WHITESPACE}"
# > length(FOO_NO_WHITESPACE)==12
如何仅删除前导空格:
FOO=' test test test '
FOO_NO_LEAD_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//')"
echo -e "FOO_NO_LEAD_SPACE='${FOO_NO_LEAD_SPACE}'"
# > FOO_NO_LEAD_SPACE='test test test '
echo -e "length(FOO_NO_LEAD_SPACE)==${#FOO_NO_LEAD_SPACE}"
# > length(FOO_NO_LEAD_SPACE)==15
如何删除尾随空格:
FOO=' test test test '
FOO_NO_TRAIL_SPACE="$(echo -e "${FOO}" | sed -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_TRAIL_SPACE='${FOO_NO_TRAIL_SPACE}'"
# > FOO_NO_TRAIL_SPACE=' test test test'
echo -e "length(FOO_NO_TRAIL_SPACE)==${#FOO_NO_TRAIL_SPACE}"
# > length(FOO_NO_TRAIL_SPACE)==15
如何删除前导和尾随空格-链种子:
FOO=' test test test '
FOO_NO_EXTERNAL_SPACE="$(echo -e "${FOO}" | sed -e 's/^[[:space:]]*//' -e 's/[[:space:]]*$//')"
echo -e "FOO_NO_EXTERNAL_SPACE='${FOO_NO_EXTERNAL_SPACE}'"
# > FOO_NO_EXTERNAL_SPACE='test test test'
echo -e "length(FOO_NO_EXTERNAL_SPACE)==${#FOO_NO_EXTERNAL_SPACE}"
# > length(FOO_NO_EXTERNAL_SPACE)==14
或者，如果你的bash支持它，你可以替换echo -e "${FOO}" | sed…有了sed…<<<${FOO}，像这样(用于尾随空格):
FOO_NO_TRAIL_SPACE="$(sed -e 's/[[:space:]]*$//' <<<${FOO})"
其他回答
#Execute this script with the string argument passed in double quotes !!
#var2 gives the string without spaces.
#$1 is the string passed in double quotes
#!/bin/bash
var2=`echo $1 | sed 's/ \+//g'`
echo $var2
一个简单的答案是:
echo " lol " | xargs
Xargs将为您做修剪。这是一个命令/程序，没有参数，返回修剪后的字符串，就这么简单!
注意:这并没有删除所有的内部空格，所以“foo bar”保持不变;它不会变成“foobar”。但是，多个空格将被压缩为单个空格，因此“foo bar”将变成“foo bar”。此外，它不会删除行尾字符。
你可以简单地使用echo进行修剪:
foo=" qsdqsd qsdqs q qs "
# Not trimmed
echo \'$foo\'
# Trim
foo=`echo $foo`
# Trimmed
echo \'$foo\'
使用AWK:
echo $var | awk '{gsub(/^ +| +$/,"")}1'
还有一个单元测试的解决方案，它从stdin中删除$IFS，并适用于任何输入分隔符(甚至$'\0'):
ltrim()
{
# Left-trim $IFS from stdin as a single line
# $1: Line separator (default NUL)
local trimmed
while IFS= read -r -d "${1-}" -u 9
do
if [ -n "${trimmed+defined}" ]
then
printf %s "$REPLY"
else
printf %s "${REPLY#"${REPLY%%[!$IFS]*}"}"
fi
printf "${1-\x00}"
trimmed=true
done 9<&0
if [[ $REPLY ]]
then
# No delimiter at last line
if [ -n "${trimmed+defined}" ]
then
printf %s "$REPLY"
else
printf %s "${REPLY#"${REPLY%%[!$IFS]*}"}"
fi
fi
}
rtrim()
{
# Right-trim $IFS from stdin as a single line
# $1: Line separator (default NUL)
local previous last
while IFS= read -r -d "${1-}" -u 9
do
if [ -n "${previous+defined}" ]
then
printf %s "$previous"
printf "${1-\x00}"
fi
previous="$REPLY"
done 9<&0
if [[ $REPLY ]]
then
# No delimiter at last line
last="$REPLY"
printf %s "$previous"
if [ -n "${previous+defined}" ]
then
printf "${1-\x00}"
fi
else
last="$previous"
fi
right_whitespace="${last##*[!$IFS]}"
printf %s "${last%$right_whitespace}"
}
trim()
{
# Trim $IFS from individual lines
# $1: Line separator (default NUL)
ltrim ${1+"$@"} | rtrim ${1+"$@"}
}
