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
我发现我需要从混乱的sdiff输出中添加一些代码来清理它:
sdiff -s column1.txt column2.txt | grep -F '<' | cut -f1 -d"<" > c12diff.txt
sed -n 1'p' c12diff.txt | sed 's/ *$//g' | tr -d '\n' | tr -d '\t'
这将删除尾随空格和其他不可见字符。
其他回答
将空格移到一个空格:
(text) | fmt -su
我见过脚本只是使用变量赋值来完成工作:
$ xyz=`echo -e 'foo \n bar'`
$ echo $xyz
foo bar
空格会自动合并和修剪。必须小心shell元字符(潜在的注入风险)。
我还建议在shell条件句中使用双引号变量替换:
if [ -n "$var" ]; then
因为变量中的-o或其他内容可能会修改测试参数。
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<
创建一个数组而不是变量，这将修剪所有的空格，制表符和换行符:
arr=( $(hg st -R "$path") )
if [[ -n "${arr[@]}" ]]; then
printf -- '%s\n' "${arr[@]}"
fi
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
