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"函数删除所有水平空白:
ltrim () {
if [[ $# -eq 0 ]]; then cat; else printf -- '%s\n' "$@"; fi | perl -pe 's/^\h+//g'
return $?
}
rtrim () {
if [[ $# -eq 0 ]]; then cat; else printf -- '%s\n' "$@"; fi | perl -pe 's/\h+$//g'
return $?
}
trim () {
ltrim "$@" | rtrim
return $?
}
其他回答
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'
赋值函数忽略前导和后面的空格，因此可以用于修剪:
$ var=`echo ' hello'`; echo $var
hello
Read已经修整了空白，所以在bash中你可以这样做:
$ read foo <<< " foo bar two spaces follow "
$ echo ".$foo."
.foo bar two spaces follow.
POSIX兼容的版本稍长一些
$ read foo << END
foo bar two spaces follow
END
$ echo ".$foo."
.foo bar two spaces follow.
这里有一个trim()函数，用于修整和规范化空白
#!/bin/bash
function trim {
echo $*
}
echo "'$(trim " one two three ")'"
# 'one two three'
还有一种使用正则表达式的变体。
#!/bin/bash
function trim {
local trimmed="$@"
if [[ "$trimmed" =~ " *([^ ].*[^ ]) *" ]]
then
trimmed=${BASH_REMATCH[1]}
fi
echo "$trimmed"
}
echo "'$(trim " one two three ")'"
# 'one two three'
Use:
trim() {
local orig="$1"
local trmd=""
while true;
do
trmd="${orig#[[:space:]]}"
trmd="${trmd%[[:space:]]}"
test "$trmd" = "$orig" && break
orig="$trmd"
done
printf -- '%s\n' "$trmd"
}
它适用于各种空格，包括换行符， 不需要修改shop。 它保留内部空白，包括换行符。
单元测试(用于手动检查):
#!/bin/bash
. trim.sh
enum() {
echo " a b c"
echo "a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a b c "
echo " a N b c "
echo "N a N b c "
echo " Na b c "
echo " a b c N "
echo " a b c N"
}
xcheck() {
local testln result
while IFS='' read testln;
do
testln=$(tr N '\n' <<<"$testln")
echo ": ~~~~~~~~~~~~~~~~~~~~~~~~~ :" >&2
result="$(trim "$testln")"
echo "testln='$testln'" >&2
echo "result='$result'" >&2
done
}
enum | xcheck
