How to print a file, excluding comments and blank lines, using grep/sed?

Question

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: 

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

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

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

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.

Answer 1

With grep:

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

On OSX / BSD systems:

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