如何使用 grep/sed 打印文件，不包括注释和空行? [英] How to print a file, excluding comments and blank lines, using grep/sed?

问题描述

我想打印一个包含一系列注释的文件，例如:

I'd like to print out a file containing a series of comments like:

    </Directory>
    ErrorLog ${APACHE_LOG_DIR}/error.log
    # Possible values include: debug, info, notice, warn, error, crit,
    # alert, emerg.
    LogLevel warn
    CustomLog ${APACHE_LOG_DIR}/ssl_access.log combined
    #   SSL Engine Switch:

本质上，文件包含多个缩进级别，其中注释以 # 符号开头.

In essence, the file contains multiple indentation levels, where a comment starts with a # symbol.

grep 应该删除空行，以及文本前有哈希符号的行(暗示这些是注释).

grep should remove blank lines, and also lines where there is a hash symbol before text (implying that these are comments).

我知道可以通过以下方式删除空行:grep -v '^$'

I know that blank lines can be deleted via: grep -v '^$'

但是，如何删除带有前导空格的行，然后是 # 符号，并仅打印出带有实际代码的行?我想在 bash 中使用 grep 和/或 sed 执行此操作.

However how can I delete lines with leading whitespace, and then a # symbol, and print out only lines with actual code? I would like to do this in bash, using grep and/or sed.

推荐答案

With grep:

grep -v '^\s*$\|^\s*\#' temp

在 OSX/BSD 系统上:

On OSX / BSD systems:

grep -Ev '^\s*$|^\s*\#' temp

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