Unix / Linux操作系统，删除线注释 [英] Unix/Linux, Delete comments from lines

问题描述

我需要删除/删除用户输入行注释而不删除任何codeS。因此，例如：

I need to delete/remove comments from a user-input line without deleting any codes. So for example:

mail -s 'text' brown < text #comments

我如何删除评论并留下code完好？
我可以删除以＃开头的行，但如果它在线路中间的某个位置开始没有。

How do I remove the comments and leave the code intact? I can delete lines that begin with #, but not if it begins somewhere in the middle of the lines.

我试过：

echo $line | sed -e 's/\

，但它不工作。任何想法，我做错了什么？

but it does not work. Any idea what I'm doing wrong?

另外，如何检测其中＃不用于开始评论例？
对于因为它们不是注释以＃结束例如报价＃和code系列。

Also, how to detect cases in which # is not used to begin a comment? For example quoted # and line of code that ends with # since they are not comments.

echo $line | sed -e  '/^#/d'

在此行中，＃，不使用作为注释，但作为code的一部分。我弄清楚，我需要检测，如果是＃引号内或不具备前的＃空白字符。我怎么离开的输出，因为它是？

In this line, the # is not used as a comment, but as part of code. I figure out that I need to detect that if # is within quotes or does not have a whitespace character before the #. How do I leave the output as it is?

推荐答案

您可以从＃使用该 awk的<删除所有到行尾/ code>

You can remove all from # to end of line using this awk

awk '{sub(/#.*$/,"")}1' file

但是，如果你的文件是这样的：

But if you have file like this:

#!/bin/bash
pidof tail #See if tail is running
if [ $? -ne 0 ] ; then  #start loop
   awk '{print " # "$8}' file >tmp # this is my code
fi # end of loop
awk -F# '{for (i=1;i<=NF;i++) print $i}' file > tmp2
a=a+1 # increment a

有没有方法可以自动删除评论不破坏一些。

There are no way you can remove the comment automatically without destroying some.

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