Bash shell 中的错误分隔符
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Bad delimiter in Bash shell
提问by Naftaly
I'm trying to iterate over few lines of bash script and echo all the non comment lines.
我正在尝试迭代几行 bash 脚本并回显所有非注释行。
Within my loop I have the following command:
在我的循环中，我有以下命令：
echo $line | cut -d" #" -f1
However the interpreter throws at me the following error:
但是，解释器向我抛出以下错误：
cut: bad delimiter
What I did wrong?
我做错了什么？
采纳答案by Warwick
As mentioned in the comments, the -dparameter to cutexpects only one character. try using nawkinstead for your example:
正如评论中提到的，-d参数 tocut只需要一个字符。尝试使用nawk代替您的示例：
echo $line | nawk 'BEGIN {FS=" #" } ; { print }'
Or to just print lines that don't begin with a " #", use grep:
或者只打印不以“#”开头的行，请使用grep：
grep -v " #" <file>
Or to print only lines that begin with a hash, or just white space and a hash, I'd use Perl:
或者只打印以散列开头的行，或者只打印空格和散列，我会使用 Perl：
perl -n -e 'unless (/(\S+)*#/) {print}' <file>
回答by John B
You could use awk:
你可以使用awk：
awk ' !~ /^#/' file
Only prints lines that do not start with a #in the first field.
只打印#第一个字段中不以 a 开头的行。
If you want to test if a string is a comment:
如果要测试字符串是否为注释：
if [[ ! $(awk ' !~ /^#/' <<<"$string") ]]; then
echo "'$string' is a comment"
else
echo "'$string' is not a comment"
fi
Or another way using bashonly:
或bash仅使用另一种方式：
if [[ ! ${string%\#*} || ${string%\#*} == +([[:space:]]) ]]; then
echo "'$string' is a comment"
else
echo "'$string' is not a comment"
fi
回答by zerodiff
Just for completeness, using sed:
只是为了完整性，使用sed：
sed '/^\S*#/d' file
This essentially says output all lines that do not start with any whitespace followed by a #. This is a modern version of sed, so older ones you might have to use something else for the whitespace match.
这实质上是说输出所有不以任何空格开头的行，后跟 a #。这是 的现代版本sed，因此较旧的版本您可能必须使用其他东西来进行空白匹配。
