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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.
688
If you want to only match space and tab, use [[:blank:]] or [ \t] .
You can use grep with -A n option to print N lines after matching lines. Using -B n option you can print N lines before matching lines. Using -C n option you can print N lines before and after matching lines.
With grep:
grep -v '^\s*$\|^\s*\#' temp On OSX / BSD systems:
grep -Ev '^\s*$|^\s*\#' temp If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
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