bash 如何使用 Sed 取消注释包含特定字符串的行？
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How to UNCOMMENT a line that contains a specific string using Sed?
提问by JNLK
The lines in the file :
文件中的行：
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2000 -j ACCEPT
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2001 -j ACCEPT
-A INPUT -m state --state NEW -m tcp -p tcp --dport 2002 -j ACCEPT
to comment out let's say the line that contains
注释掉让我们说包含的行
2001
i can simply run this SED command:
我可以简单地运行这个 SED 命令：
sed -i '/ 2001 /s/^/#/' file
but now how do i revert back ?
但现在我如何恢复？
as in uncomment that same line ?
就像取消注释同一行一样？
i tried
我试过
sed -i '/ 2001 /s/^//' file
that does not work.
这是行不通的。
采纳答案by Avinash Raj
Try this sedcommand,
试试这个sed命令，
sed -i '/^#.* 2001 /s/^#//' file
回答by JNLK
Yes, to comment line containing specific string with sed, simply do:
是的，要使用 sed 注释包含特定字符串的行，只需执行以下操作：
sed -i '/<pattern>/s/^/#/g' file
And to uncomment it:
并取消注释：
sed -i '/<pattern>/s/^#//g' file
In your case:
在你的情况下：
sed -i '/2001/s/^/#/g' file (to comment out)
sed -i '/2001/s/^#//g' file (to uncomment)
Option "g" at the end means global change. If you want to change only a single instance of pattern, just skip this.
最后的选项“g”表示全局变化。如果您只想更改模式的单个实例，请跳过此步骤。
回答by mklement0
To complement @Avinash Raj's helpful answerwith a more generic, POSIX-compliantsolution.
用更通用的、符合 POSIX 的解决方案来补充@Avinash Raj 的有用答案。
- Togglescommenting of linesthat match a specifiable stringthat must occur as a separate wordanywhere on the line.
- The comment character(string) is also specifiable.
- 切换与指定字符串匹配的行的注释，该字符串必须作为单独的单词出现在该行的任何位置。
- 该注释字符（串），也可指定。
Note that the solution is awk-based, because a robust portablesolution with sedis virtually impossible due to the limitations of POSIX' basicregular expressions.
请注意，该解决方案是awk-based，因为由于 POSIX基本正则表达式的限制，几乎不可能实现强大的可移植解决方案。sed
awk -v commentId='#' -v word='2001' '
sed -i "" "/.*#.*d\/docker-php-ext-xdebug\.ini.*/s/^#//g" docker-compose.yml
~ "(^|[[:punct:][:space:]])" word "($|[[:punct:][:space:]])" {
if (match(umask 027
TMOUT=600
, "^[[:space:]]*" commentId))
# backup file (because we should always do this)
cp /etc/bash.bashrc /etc/bash.bashrc.$(date '+%Y-%m-%d,%H:%M:%S')
# original: TMOUT=600 , result :# TMOUT=600
sed -i '/[^#]/ s/\(^TMOUT=600.*$\)/#\ /' /etc/bash.bashrc
# original # TMOUT=600 ,result :TMOUT=600
sed -i '/^#.*TMOUT=600.*$/s/^#\ //' /etc/bash.bashrc
= substr(##代码##, RSTART + RLENGTH)
else
##代码## = commentId ##代码##
}
{ print }
' file > tmpfile.$$ && mv tmpfile.$$ file
(^|[[:punct:][:space:]])and($|[[:punct:][:space:]])are the POSIX extended regex equivalents of the\<and\>word-boundary assertions known from other regex dialects.- Whitespace afterthe comment char is preserved, but not before it.
- When prepending the comment char to a line, it is directly prepended, without whitespace.
- Thus, if you only toggle comments with thissolution, all whitespace is preserved.
- POSIX
awkdoesn't offer in-place updating (neither does POSIXsed, incidentally), hence the output is first captured in a temporary file and that file then replaces the original on success.
(^|[[:punct:][:space:]])和($|[[:punct:][:space:]])是其他正则表达式方言中已知的\<和\>词边界断言的 POSIX 扩展正则表达式等价物。- 保留注释字符后的空白，但不保留它之前的空白。
- 当将注释字符添加到一行时，它是直接添加的，没有空格。
- 因此，如果您仅使用此解决方案切换注释，则会保留所有空格。
- POSIX
awk不提供就地更新（sed顺便说一句，POSIX也不提供），因此输出首先在临时文件中捕获，然后该文件在成功时替换原始文件。
回答by max4ever
For mac, which doesn't support standard sed parameters, this would remove the hashtag
对于不支持标准 sed 参数的 mac，这将删除主题标签
##代码##回答by Mike Q
Quick example of how to comment and uncomment a line in a file.
如何注释和取消注释文件中的行的快速示例。
Sample file :
示例文件：
##代码##Lets now backup the file (just for laughs) and comment out and un comment:
现在让我们备份文件（只是为了笑）并注释掉并取消注释：
##代码##